What is a ^ b and (a & b) << 1?2019 Community Moderator ElectionWhat is the best way to add two numbers without using the + operator?Adding two numbers without + operator (Clarification)adds two numbers without using + or any arithmetic operatorsAdding two numbers without using the addition operatorWhat is the most efficient way to deep clone an object in JavaScript?What is the preferred syntax for defining enums in JavaScript?What is the scope of variables in JavaScript?What is the !! (not not) operator in JavaScript?What is the JavaScript version of sleep()?What does “use strict” do in JavaScript, and what is the reasoning behind it?What is the 'new' keyword in JavaScript?What is the difference between call and apply?What is JSONP, and why was it created?What is the difference between Bower and npm?

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What is a ^ b and (a & b)



2019 Community Moderator ElectionWhat is the best way to add two numbers without using the + operator?Adding two numbers without + operator (Clarification)adds two numbers without using + or any arithmetic operatorsAdding two numbers without using the addition operatorWhat is the most efficient way to deep clone an object in JavaScript?What is the preferred syntax for defining enums in JavaScript?What is the scope of variables in JavaScript?What is the !! (not not) operator in JavaScript?What is the JavaScript version of sleep()?What does “use strict” do in JavaScript, and what is the reasoning behind it?What is the 'new' keyword in JavaScript?What is the difference between call and apply?What is JSONP, and why was it created?What is the difference between Bower and npm?










20















I was doing this question in leetcode.



Request:




Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.




I can't understand the solution it gave



Could someone explain how this getSum function works?



Here is the answer in JS:






var getSum=function(a,b) 
const Sum = a^b; //I can't understand how those two line's code can
const carry = (a & b) << 1; //get the sum
if(!carry)
return Sum

return getSum(Sum,carry);
;
console.log(getSum(5,1));












share|improve this question



















  • 1





    Please edit your question to clarify what it is that you don't understand about those lines. Are you unfamiliar with what the ^, & and << operators do in JavaScript? Or are you just confused about how they can be used to calculate the sum of two numbers? "I can't understand it" is not a good question.

    – Ilmari Karonen
    2 days ago






  • 1





    Possible duplicate of Adding two numbers without + operator (Clarification)

    – Iłya Bursov
    yesterday











  • @Ilmari Karonen Ok...I'll edit this question

    – Jacky
    yesterday















20















I was doing this question in leetcode.



Request:




Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.




I can't understand the solution it gave



Could someone explain how this getSum function works?



Here is the answer in JS:






var getSum=function(a,b) 
const Sum = a^b; //I can't understand how those two line's code can
const carry = (a & b) << 1; //get the sum
if(!carry)
return Sum

return getSum(Sum,carry);
;
console.log(getSum(5,1));












share|improve this question



















  • 1





    Please edit your question to clarify what it is that you don't understand about those lines. Are you unfamiliar with what the ^, & and << operators do in JavaScript? Or are you just confused about how they can be used to calculate the sum of two numbers? "I can't understand it" is not a good question.

    – Ilmari Karonen
    2 days ago






  • 1





    Possible duplicate of Adding two numbers without + operator (Clarification)

    – Iłya Bursov
    yesterday











  • @Ilmari Karonen Ok...I'll edit this question

    – Jacky
    yesterday













20












20








20


4






I was doing this question in leetcode.



Request:




Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.




I can't understand the solution it gave



Could someone explain how this getSum function works?



Here is the answer in JS:






var getSum=function(a,b) 
const Sum = a^b; //I can't understand how those two line's code can
const carry = (a & b) << 1; //get the sum
if(!carry)
return Sum

return getSum(Sum,carry);
;
console.log(getSum(5,1));












share|improve this question
















I was doing this question in leetcode.



Request:




Calculate the sum of two integers a and b, but you are not allowed to use the operator + and -.




I can't understand the solution it gave



Could someone explain how this getSum function works?



Here is the answer in JS:






var getSum=function(a,b) 
const Sum = a^b; //I can't understand how those two line's code can
const carry = (a & b) << 1; //get the sum
if(!carry)
return Sum

return getSum(Sum,carry);
;
console.log(getSum(5,1));








var getSum=function(a,b) 
const Sum = a^b; //I can't understand how those two line's code can
const carry = (a & b) << 1; //get the sum
if(!carry)
return Sum

return getSum(Sum,carry);
;
console.log(getSum(5,1));





var getSum=function(a,b) 
const Sum = a^b; //I can't understand how those two line's code can
const carry = (a & b) << 1; //get the sum
if(!carry)
return Sum

return getSum(Sum,carry);
;
console.log(getSum(5,1));






javascript bitwise-operators






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday









flppv

1,527726




1,527726










asked 2 days ago









JackyJacky

26210




26210







  • 1





    Please edit your question to clarify what it is that you don't understand about those lines. Are you unfamiliar with what the ^, & and << operators do in JavaScript? Or are you just confused about how they can be used to calculate the sum of two numbers? "I can't understand it" is not a good question.

    – Ilmari Karonen
    2 days ago






  • 1





    Possible duplicate of Adding two numbers without + operator (Clarification)

    – Iłya Bursov
    yesterday











  • @Ilmari Karonen Ok...I'll edit this question

    – Jacky
    yesterday












  • 1





    Please edit your question to clarify what it is that you don't understand about those lines. Are you unfamiliar with what the ^, & and << operators do in JavaScript? Or are you just confused about how they can be used to calculate the sum of two numbers? "I can't understand it" is not a good question.

    – Ilmari Karonen
    2 days ago






  • 1





    Possible duplicate of Adding two numbers without + operator (Clarification)

    – Iłya Bursov
    yesterday











  • @Ilmari Karonen Ok...I'll edit this question

    – Jacky
    yesterday







1




1





Please edit your question to clarify what it is that you don't understand about those lines. Are you unfamiliar with what the ^, & and << operators do in JavaScript? Or are you just confused about how they can be used to calculate the sum of two numbers? "I can't understand it" is not a good question.

– Ilmari Karonen
2 days ago





Please edit your question to clarify what it is that you don't understand about those lines. Are you unfamiliar with what the ^, & and << operators do in JavaScript? Or are you just confused about how they can be used to calculate the sum of two numbers? "I can't understand it" is not a good question.

– Ilmari Karonen
2 days ago




1




1





Possible duplicate of Adding two numbers without + operator (Clarification)

– Iłya Bursov
yesterday





Possible duplicate of Adding two numbers without + operator (Clarification)

– Iłya Bursov
yesterday













@Ilmari Karonen Ok...I'll edit this question

– Jacky
yesterday





@Ilmari Karonen Ok...I'll edit this question

– Jacky
yesterday












4 Answers
4






active

oldest

votes


















22














Let's learn by example. Imagine that a = 3 and b = 5



In binary notation they are a = 0011 and b = 0101



XOR:
a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11



So when we're doing a^b result will be 0110.



AND + SHIFT



a&b performs logical AND operation. It returns 1 only when a = b = 1.



In our case the result is 0001



<< shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).



Next iterations:



Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)



Now a^b = 0100 and (a&b)<<1 = 0100



Repeating again.



Now a^b = 0000 and (a&b)<<1 = 1000



And again.



Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.



Everything worked fine since 3+5=8






share|improve this answer




















  • 2





    This only seems to discuss that particular case, not how the algorithm works in general. How are we to know from this that it works for all possible inputs?

    – ilkkachu
    yesterday











  • @ilkkachu you can try different input values, it will work with maybe different number of iterations. My goal was to explain the algorithm by example.

    – flppv
    yesterday


















31














It's basically replicating the half-adder



Adding 2 bits A and B produces 2 outputs: a sum and a carry bit like below



╔═══════╤═════════════╗
║ Input │ Output ║
╠═══╤═══╪═══════╤═════╣
║ A │ B │ carry │ sum ║
╟───┼───┼───────┼─────╢
║ 0 │ 0 │ 0 │ 0 ║
╟───┼───┼───────┼─────╢
║ 1 │ 0 │ 0 │ 1 ║
╟───┼───┼───────┼─────╢
║ 0 │ 1 │ 0 │ 1 ║
╟───┼───┼───────┼─────╢
║ 1 │ 1 │ 1 │ 0 ║
╚═══╧═══╧═══════╧═════╝


From the table we get the logic for the outputs: carry = A and B, sum = A xor B





XOR is also called a carry-less add operator, and represented by ⊕ with the + symbol inside



So the snippet above is working like this



const Sum=a^b; // sum = a xor b = a ⊕ b
const carry=(a&b)<<1; // carry = 2*(a and b), since we carry to the next bit
if(!carry)
return Sum; // no carry, so sum + carry = sum

return getSum(Sum,carry); // a + b = sum + carry


So a^b adds each bit in a and b simultaneously, leaving the non-carry sum of a and b in Sum. Then we have to add carry to the carry-less sum to get the final result, since we have only a half-adder instead of a full-adder which does a + b = a ⊕ b + carry



See also



  • Adding two numbers without + operator (Clarification)

  • What is the best way to add two numbers without using the + operator?

  • adds two numbers without using + or any arithmetic operators

  • Adding two numbers without using the addition operator





share|improve this answer

























  • If I'm not mistaken, it would be more accurate to say it replicates full adder, because of the function's recursive call to itself - one xor plus and stage leading to another.

    – Sergiy Kolodyazhnyy
    2 days ago











  • @SergiyKolodyazhnyy here the input has only a and b which is the half adder. The full adder capability is achieved by the recursive call which gives you a + b + 2*carry. A real full adder takes 3 inputs a + b + carryIn = sum + 2*carryOut

    – phuclv
    2 days ago






  • 2





    Might be worth pointing out that it's a half-adder for each bit simultaneously, unlike the truth table that only shows a single bit position.

    – ilkkachu
    yesterday


















2














 int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
int carry = (p & q) << 1; // Left Shift, 1+1=2
if (carry != 0)
return getSum(result, carry);

return result;


Start By p=5,q=6. Then the XOR would be,



0101
0110
------
0011


So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,



0101
0110
-------
0100


We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get



 0100<<1=1000


So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.



getSum(3, 8);


So, doing the first XORing we get,



0011
1000
-------
1011


The XORing this time yielded in 11 (1011 binary),so we perform the AND now,



0011
1000
-------
0000


We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).



Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.






share|improve this answer
































    0














    ^ is XOR, a bitwise operation. On a single bit, the rules are 0 ^ 0 = 0, 0 ^ 1 = 1, 1 ^ 0 = 0, and 1 ^ 1 = 0, and you simply extend perform it on corresponding bits when dealing with multi-bit values. The name is short for "exclusive or", and comes from the fact that A ^ B is 1 if and only if either A or B is 1, not both. But, it's more interesting to talk about its other name, ⊕. ⊕ is + but slightly different. You'll notice that the rules for ⊕ are similar to the rules for addition: 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, and 1 + 1 = 10. ⊕ is +, except 1 ⊕ 1 = 0; that is, ⊕ is +, except without carrying. This holds for multiple bits: 011 + 001 = 100, because you carry a 1 out of the ones place into the twos place, and then carry a 1 again into the fours place. Then, 011 ⊕ 001 = 010, because you just don't carry.



    Now, when doing real addition, when do you carry? In binary, the answer is very simple: you carry a 1 into the next place when there are two 1s in a given place. This is easily understood as a bitwise AND, &. 1 & 1 = 1, and 0 otherwise. For 011 + 001, addition without carrying gives 011 ⊕ 001 = 010, and we can tell we need to carry a 1 out of the ones place because 011 & 001 = 001. The shifting in (a & b) << 1 turns a number "where do I need to carry from?" into "where do I need to add carries?": (011 & 001) << 1 = 010; I need to add a carry bit in the twos place.



    So, in getSum, we want to know a + b. We compute the addition without carrying with a ^ b, and we find where we need to add carry bits with (a & b) << 1. Now, we just need to add those two together. Well, we already have a function for adding numbers together; it's called getSum. So, we basically just write function getSum(a, b) return getSum(a ^ b, (a & b) << 1); , except we make sure to short-circuit if there is nothing to carry, saving us from infinite recursion.






    share|improve this answer






















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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      22














      Let's learn by example. Imagine that a = 3 and b = 5



      In binary notation they are a = 0011 and b = 0101



      XOR:
      a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11



      So when we're doing a^b result will be 0110.



      AND + SHIFT



      a&b performs logical AND operation. It returns 1 only when a = b = 1.



      In our case the result is 0001



      << shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).



      Next iterations:



      Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)



      Now a^b = 0100 and (a&b)<<1 = 0100



      Repeating again.



      Now a^b = 0000 and (a&b)<<1 = 1000



      And again.



      Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.



      Everything worked fine since 3+5=8






      share|improve this answer




















      • 2





        This only seems to discuss that particular case, not how the algorithm works in general. How are we to know from this that it works for all possible inputs?

        – ilkkachu
        yesterday











      • @ilkkachu you can try different input values, it will work with maybe different number of iterations. My goal was to explain the algorithm by example.

        – flppv
        yesterday















      22














      Let's learn by example. Imagine that a = 3 and b = 5



      In binary notation they are a = 0011 and b = 0101



      XOR:
      a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11



      So when we're doing a^b result will be 0110.



      AND + SHIFT



      a&b performs logical AND operation. It returns 1 only when a = b = 1.



      In our case the result is 0001



      << shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).



      Next iterations:



      Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)



      Now a^b = 0100 and (a&b)<<1 = 0100



      Repeating again.



      Now a^b = 0000 and (a&b)<<1 = 1000



      And again.



      Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.



      Everything worked fine since 3+5=8






      share|improve this answer




















      • 2





        This only seems to discuss that particular case, not how the algorithm works in general. How are we to know from this that it works for all possible inputs?

        – ilkkachu
        yesterday











      • @ilkkachu you can try different input values, it will work with maybe different number of iterations. My goal was to explain the algorithm by example.

        – flppv
        yesterday













      22












      22








      22







      Let's learn by example. Imagine that a = 3 and b = 5



      In binary notation they are a = 0011 and b = 0101



      XOR:
      a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11



      So when we're doing a^b result will be 0110.



      AND + SHIFT



      a&b performs logical AND operation. It returns 1 only when a = b = 1.



      In our case the result is 0001



      << shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).



      Next iterations:



      Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)



      Now a^b = 0100 and (a&b)<<1 = 0100



      Repeating again.



      Now a^b = 0000 and (a&b)<<1 = 1000



      And again.



      Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.



      Everything worked fine since 3+5=8






      share|improve this answer















      Let's learn by example. Imagine that a = 3 and b = 5



      In binary notation they are a = 0011 and b = 0101



      XOR:
      a^b is XOR operator. When compare two bits it returns 0 if they are same and 1 if they are different. 01^10 => 11



      So when we're doing a^b result will be 0110.



      AND + SHIFT



      a&b performs logical AND operation. It returns 1 only when a = b = 1.



      In our case the result is 0001



      << shifts it(adds 0 on the right side) and result became 0010 which sets carry variable true. (only 0000 will be false).



      Next iterations:



      Everything repeats but now a = 0110 and b = 0010 (Sum and carry from last execution)



      Now a^b = 0100 and (a&b)<<1 = 0100



      Repeating again.



      Now a^b = 0000 and (a&b)<<1 = 1000



      And again.



      Now a^b = 1000 and (a&b)<<1 = 0000. Now carry is finally false. And we're returning 1000 which is decimal 8.



      Everything worked fine since 3+5=8







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited yesterday

























      answered 2 days ago









      flppvflppv

      1,527726




      1,527726







      • 2





        This only seems to discuss that particular case, not how the algorithm works in general. How are we to know from this that it works for all possible inputs?

        – ilkkachu
        yesterday











      • @ilkkachu you can try different input values, it will work with maybe different number of iterations. My goal was to explain the algorithm by example.

        – flppv
        yesterday












      • 2





        This only seems to discuss that particular case, not how the algorithm works in general. How are we to know from this that it works for all possible inputs?

        – ilkkachu
        yesterday











      • @ilkkachu you can try different input values, it will work with maybe different number of iterations. My goal was to explain the algorithm by example.

        – flppv
        yesterday







      2




      2





      This only seems to discuss that particular case, not how the algorithm works in general. How are we to know from this that it works for all possible inputs?

      – ilkkachu
      yesterday





      This only seems to discuss that particular case, not how the algorithm works in general. How are we to know from this that it works for all possible inputs?

      – ilkkachu
      yesterday













      @ilkkachu you can try different input values, it will work with maybe different number of iterations. My goal was to explain the algorithm by example.

      – flppv
      yesterday





      @ilkkachu you can try different input values, it will work with maybe different number of iterations. My goal was to explain the algorithm by example.

      – flppv
      yesterday













      31














      It's basically replicating the half-adder



      Adding 2 bits A and B produces 2 outputs: a sum and a carry bit like below



      ╔═══════╤═════════════╗
      ║ Input │ Output ║
      ╠═══╤═══╪═══════╤═════╣
      ║ A │ B │ carry │ sum ║
      ╟───┼───┼───────┼─────╢
      ║ 0 │ 0 │ 0 │ 0 ║
      ╟───┼───┼───────┼─────╢
      ║ 1 │ 0 │ 0 │ 1 ║
      ╟───┼───┼───────┼─────╢
      ║ 0 │ 1 │ 0 │ 1 ║
      ╟───┼───┼───────┼─────╢
      ║ 1 │ 1 │ 1 │ 0 ║
      ╚═══╧═══╧═══════╧═════╝


      From the table we get the logic for the outputs: carry = A and B, sum = A xor B





      XOR is also called a carry-less add operator, and represented by ⊕ with the + symbol inside



      So the snippet above is working like this



      const Sum=a^b; // sum = a xor b = a ⊕ b
      const carry=(a&b)<<1; // carry = 2*(a and b), since we carry to the next bit
      if(!carry)
      return Sum; // no carry, so sum + carry = sum

      return getSum(Sum,carry); // a + b = sum + carry


      So a^b adds each bit in a and b simultaneously, leaving the non-carry sum of a and b in Sum. Then we have to add carry to the carry-less sum to get the final result, since we have only a half-adder instead of a full-adder which does a + b = a ⊕ b + carry



      See also



      • Adding two numbers without + operator (Clarification)

      • What is the best way to add two numbers without using the + operator?

      • adds two numbers without using + or any arithmetic operators

      • Adding two numbers without using the addition operator





      share|improve this answer

























      • If I'm not mistaken, it would be more accurate to say it replicates full adder, because of the function's recursive call to itself - one xor plus and stage leading to another.

        – Sergiy Kolodyazhnyy
        2 days ago











      • @SergiyKolodyazhnyy here the input has only a and b which is the half adder. The full adder capability is achieved by the recursive call which gives you a + b + 2*carry. A real full adder takes 3 inputs a + b + carryIn = sum + 2*carryOut

        – phuclv
        2 days ago






      • 2





        Might be worth pointing out that it's a half-adder for each bit simultaneously, unlike the truth table that only shows a single bit position.

        – ilkkachu
        yesterday















      31














      It's basically replicating the half-adder



      Adding 2 bits A and B produces 2 outputs: a sum and a carry bit like below



      ╔═══════╤═════════════╗
      ║ Input │ Output ║
      ╠═══╤═══╪═══════╤═════╣
      ║ A │ B │ carry │ sum ║
      ╟───┼───┼───────┼─────╢
      ║ 0 │ 0 │ 0 │ 0 ║
      ╟───┼───┼───────┼─────╢
      ║ 1 │ 0 │ 0 │ 1 ║
      ╟───┼───┼───────┼─────╢
      ║ 0 │ 1 │ 0 │ 1 ║
      ╟───┼───┼───────┼─────╢
      ║ 1 │ 1 │ 1 │ 0 ║
      ╚═══╧═══╧═══════╧═════╝


      From the table we get the logic for the outputs: carry = A and B, sum = A xor B





      XOR is also called a carry-less add operator, and represented by ⊕ with the + symbol inside



      So the snippet above is working like this



      const Sum=a^b; // sum = a xor b = a ⊕ b
      const carry=(a&b)<<1; // carry = 2*(a and b), since we carry to the next bit
      if(!carry)
      return Sum; // no carry, so sum + carry = sum

      return getSum(Sum,carry); // a + b = sum + carry


      So a^b adds each bit in a and b simultaneously, leaving the non-carry sum of a and b in Sum. Then we have to add carry to the carry-less sum to get the final result, since we have only a half-adder instead of a full-adder which does a + b = a ⊕ b + carry



      See also



      • Adding two numbers without + operator (Clarification)

      • What is the best way to add two numbers without using the + operator?

      • adds two numbers without using + or any arithmetic operators

      • Adding two numbers without using the addition operator





      share|improve this answer

























      • If I'm not mistaken, it would be more accurate to say it replicates full adder, because of the function's recursive call to itself - one xor plus and stage leading to another.

        – Sergiy Kolodyazhnyy
        2 days ago











      • @SergiyKolodyazhnyy here the input has only a and b which is the half adder. The full adder capability is achieved by the recursive call which gives you a + b + 2*carry. A real full adder takes 3 inputs a + b + carryIn = sum + 2*carryOut

        – phuclv
        2 days ago






      • 2





        Might be worth pointing out that it's a half-adder for each bit simultaneously, unlike the truth table that only shows a single bit position.

        – ilkkachu
        yesterday













      31












      31








      31







      It's basically replicating the half-adder



      Adding 2 bits A and B produces 2 outputs: a sum and a carry bit like below



      ╔═══════╤═════════════╗
      ║ Input │ Output ║
      ╠═══╤═══╪═══════╤═════╣
      ║ A │ B │ carry │ sum ║
      ╟───┼───┼───────┼─────╢
      ║ 0 │ 0 │ 0 │ 0 ║
      ╟───┼───┼───────┼─────╢
      ║ 1 │ 0 │ 0 │ 1 ║
      ╟───┼───┼───────┼─────╢
      ║ 0 │ 1 │ 0 │ 1 ║
      ╟───┼───┼───────┼─────╢
      ║ 1 │ 1 │ 1 │ 0 ║
      ╚═══╧═══╧═══════╧═════╝


      From the table we get the logic for the outputs: carry = A and B, sum = A xor B





      XOR is also called a carry-less add operator, and represented by ⊕ with the + symbol inside



      So the snippet above is working like this



      const Sum=a^b; // sum = a xor b = a ⊕ b
      const carry=(a&b)<<1; // carry = 2*(a and b), since we carry to the next bit
      if(!carry)
      return Sum; // no carry, so sum + carry = sum

      return getSum(Sum,carry); // a + b = sum + carry


      So a^b adds each bit in a and b simultaneously, leaving the non-carry sum of a and b in Sum. Then we have to add carry to the carry-less sum to get the final result, since we have only a half-adder instead of a full-adder which does a + b = a ⊕ b + carry



      See also



      • Adding two numbers without + operator (Clarification)

      • What is the best way to add two numbers without using the + operator?

      • adds two numbers without using + or any arithmetic operators

      • Adding two numbers without using the addition operator





      share|improve this answer















      It's basically replicating the half-adder



      Adding 2 bits A and B produces 2 outputs: a sum and a carry bit like below



      ╔═══════╤═════════════╗
      ║ Input │ Output ║
      ╠═══╤═══╪═══════╤═════╣
      ║ A │ B │ carry │ sum ║
      ╟───┼───┼───────┼─────╢
      ║ 0 │ 0 │ 0 │ 0 ║
      ╟───┼───┼───────┼─────╢
      ║ 1 │ 0 │ 0 │ 1 ║
      ╟───┼───┼───────┼─────╢
      ║ 0 │ 1 │ 0 │ 1 ║
      ╟───┼───┼───────┼─────╢
      ║ 1 │ 1 │ 1 │ 0 ║
      ╚═══╧═══╧═══════╧═════╝


      From the table we get the logic for the outputs: carry = A and B, sum = A xor B





      XOR is also called a carry-less add operator, and represented by ⊕ with the + symbol inside



      So the snippet above is working like this



      const Sum=a^b; // sum = a xor b = a ⊕ b
      const carry=(a&b)<<1; // carry = 2*(a and b), since we carry to the next bit
      if(!carry)
      return Sum; // no carry, so sum + carry = sum

      return getSum(Sum,carry); // a + b = sum + carry


      So a^b adds each bit in a and b simultaneously, leaving the non-carry sum of a and b in Sum. Then we have to add carry to the carry-less sum to get the final result, since we have only a half-adder instead of a full-adder which does a + b = a ⊕ b + carry



      See also



      • Adding two numbers without + operator (Clarification)

      • What is the best way to add two numbers without using the + operator?

      • adds two numbers without using + or any arithmetic operators

      • Adding two numbers without using the addition operator






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited yesterday

























      answered 2 days ago









      phuclvphuclv

      15.6k855230




      15.6k855230












      • If I'm not mistaken, it would be more accurate to say it replicates full adder, because of the function's recursive call to itself - one xor plus and stage leading to another.

        – Sergiy Kolodyazhnyy
        2 days ago











      • @SergiyKolodyazhnyy here the input has only a and b which is the half adder. The full adder capability is achieved by the recursive call which gives you a + b + 2*carry. A real full adder takes 3 inputs a + b + carryIn = sum + 2*carryOut

        – phuclv
        2 days ago






      • 2





        Might be worth pointing out that it's a half-adder for each bit simultaneously, unlike the truth table that only shows a single bit position.

        – ilkkachu
        yesterday

















      • If I'm not mistaken, it would be more accurate to say it replicates full adder, because of the function's recursive call to itself - one xor plus and stage leading to another.

        – Sergiy Kolodyazhnyy
        2 days ago











      • @SergiyKolodyazhnyy here the input has only a and b which is the half adder. The full adder capability is achieved by the recursive call which gives you a + b + 2*carry. A real full adder takes 3 inputs a + b + carryIn = sum + 2*carryOut

        – phuclv
        2 days ago






      • 2





        Might be worth pointing out that it's a half-adder for each bit simultaneously, unlike the truth table that only shows a single bit position.

        – ilkkachu
        yesterday
















      If I'm not mistaken, it would be more accurate to say it replicates full adder, because of the function's recursive call to itself - one xor plus and stage leading to another.

      – Sergiy Kolodyazhnyy
      2 days ago





      If I'm not mistaken, it would be more accurate to say it replicates full adder, because of the function's recursive call to itself - one xor plus and stage leading to another.

      – Sergiy Kolodyazhnyy
      2 days ago













      @SergiyKolodyazhnyy here the input has only a and b which is the half adder. The full adder capability is achieved by the recursive call which gives you a + b + 2*carry. A real full adder takes 3 inputs a + b + carryIn = sum + 2*carryOut

      – phuclv
      2 days ago





      @SergiyKolodyazhnyy here the input has only a and b which is the half adder. The full adder capability is achieved by the recursive call which gives you a + b + 2*carry. A real full adder takes 3 inputs a + b + carryIn = sum + 2*carryOut

      – phuclv
      2 days ago




      2




      2





      Might be worth pointing out that it's a half-adder for each bit simultaneously, unlike the truth table that only shows a single bit position.

      – ilkkachu
      yesterday





      Might be worth pointing out that it's a half-adder for each bit simultaneously, unlike the truth table that only shows a single bit position.

      – ilkkachu
      yesterday











      2














       int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
      int carry = (p & q) << 1; // Left Shift, 1+1=2
      if (carry != 0)
      return getSum(result, carry);

      return result;


      Start By p=5,q=6. Then the XOR would be,



      0101
      0110
      ------
      0011


      So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,



      0101
      0110
      -------
      0100


      We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get



       0100<<1=1000


      So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.



      getSum(3, 8);


      So, doing the first XORing we get,



      0011
      1000
      -------
      1011


      The XORing this time yielded in 11 (1011 binary),so we perform the AND now,



      0011
      1000
      -------
      0000


      We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
      As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).



      Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.






      share|improve this answer





























        2














         int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
        int carry = (p & q) << 1; // Left Shift, 1+1=2
        if (carry != 0)
        return getSum(result, carry);

        return result;


        Start By p=5,q=6. Then the XOR would be,



        0101
        0110
        ------
        0011


        So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,



        0101
        0110
        -------
        0100


        We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get



         0100<<1=1000


        So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.



        getSum(3, 8);


        So, doing the first XORing we get,



        0011
        1000
        -------
        1011


        The XORing this time yielded in 11 (1011 binary),so we perform the AND now,



        0011
        1000
        -------
        0000


        We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
        As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).



        Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.






        share|improve this answer



























          2












          2








          2







           int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
          int carry = (p & q) << 1; // Left Shift, 1+1=2
          if (carry != 0)
          return getSum(result, carry);

          return result;


          Start By p=5,q=6. Then the XOR would be,



          0101
          0110
          ------
          0011


          So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,



          0101
          0110
          -------
          0100


          We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get



           0100<<1=1000


          So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.



          getSum(3, 8);


          So, doing the first XORing we get,



          0011
          1000
          -------
          1011


          The XORing this time yielded in 11 (1011 binary),so we perform the AND now,



          0011
          1000
          -------
          0000


          We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
          As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).



          Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.






          share|improve this answer















           int result = p ^ q; // XOR Operator, + without carry 0+0=0, 0+1=1+0=1, 1+1=0
          int carry = (p & q) << 1; // Left Shift, 1+1=2
          if (carry != 0)
          return getSum(result, carry);

          return result;


          Start By p=5,q=6. Then the XOR would be,



          0101
          0110
          ------
          0011


          So, XORing results in (0011) which is actually 3 in decimal. Then ANDing p and q we get,



          0101
          0110
          -------
          0100


          We get 4 (100 in binary) by ANDing 5 & 6, now if we left shift this value by 1, we get



           0100<<1=1000


          So we get 8 (1000 in binary) after first recursion.As the result (carry variable) isnt zero, lets recursion again by xor value and carry value.



          getSum(3, 8);


          So, doing the first XORing we get,



          0011
          1000
          -------
          1011


          The XORing this time yielded in 11 (1011 binary),so we perform the AND now,



          0011
          1000
          -------
          0000


          We get all ZERO for ANDing 3 and 8, so this time the left shift operator also results in ZERO, as we have no 1 here which may give us a value by left shifing zeroes.
          As the carry variable is now Zero, we come to the end of recursion and the XORed value will be the Sum, which is 11 (1011 in Binary).



          Hope you get the working of the procedure. You can learn more by learning bitwise operation, as its the way the machine do the arithmatic operations.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 2 days ago

























          answered 2 days ago









          Ayan_84Ayan_84

          530513




          530513





















              0














              ^ is XOR, a bitwise operation. On a single bit, the rules are 0 ^ 0 = 0, 0 ^ 1 = 1, 1 ^ 0 = 0, and 1 ^ 1 = 0, and you simply extend perform it on corresponding bits when dealing with multi-bit values. The name is short for "exclusive or", and comes from the fact that A ^ B is 1 if and only if either A or B is 1, not both. But, it's more interesting to talk about its other name, ⊕. ⊕ is + but slightly different. You'll notice that the rules for ⊕ are similar to the rules for addition: 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, and 1 + 1 = 10. ⊕ is +, except 1 ⊕ 1 = 0; that is, ⊕ is +, except without carrying. This holds for multiple bits: 011 + 001 = 100, because you carry a 1 out of the ones place into the twos place, and then carry a 1 again into the fours place. Then, 011 ⊕ 001 = 010, because you just don't carry.



              Now, when doing real addition, when do you carry? In binary, the answer is very simple: you carry a 1 into the next place when there are two 1s in a given place. This is easily understood as a bitwise AND, &. 1 & 1 = 1, and 0 otherwise. For 011 + 001, addition without carrying gives 011 ⊕ 001 = 010, and we can tell we need to carry a 1 out of the ones place because 011 & 001 = 001. The shifting in (a & b) << 1 turns a number "where do I need to carry from?" into "where do I need to add carries?": (011 & 001) << 1 = 010; I need to add a carry bit in the twos place.



              So, in getSum, we want to know a + b. We compute the addition without carrying with a ^ b, and we find where we need to add carry bits with (a & b) << 1. Now, we just need to add those two together. Well, we already have a function for adding numbers together; it's called getSum. So, we basically just write function getSum(a, b) return getSum(a ^ b, (a & b) << 1); , except we make sure to short-circuit if there is nothing to carry, saving us from infinite recursion.






              share|improve this answer



























                0














                ^ is XOR, a bitwise operation. On a single bit, the rules are 0 ^ 0 = 0, 0 ^ 1 = 1, 1 ^ 0 = 0, and 1 ^ 1 = 0, and you simply extend perform it on corresponding bits when dealing with multi-bit values. The name is short for "exclusive or", and comes from the fact that A ^ B is 1 if and only if either A or B is 1, not both. But, it's more interesting to talk about its other name, ⊕. ⊕ is + but slightly different. You'll notice that the rules for ⊕ are similar to the rules for addition: 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, and 1 + 1 = 10. ⊕ is +, except 1 ⊕ 1 = 0; that is, ⊕ is +, except without carrying. This holds for multiple bits: 011 + 001 = 100, because you carry a 1 out of the ones place into the twos place, and then carry a 1 again into the fours place. Then, 011 ⊕ 001 = 010, because you just don't carry.



                Now, when doing real addition, when do you carry? In binary, the answer is very simple: you carry a 1 into the next place when there are two 1s in a given place. This is easily understood as a bitwise AND, &. 1 & 1 = 1, and 0 otherwise. For 011 + 001, addition without carrying gives 011 ⊕ 001 = 010, and we can tell we need to carry a 1 out of the ones place because 011 & 001 = 001. The shifting in (a & b) << 1 turns a number "where do I need to carry from?" into "where do I need to add carries?": (011 & 001) << 1 = 010; I need to add a carry bit in the twos place.



                So, in getSum, we want to know a + b. We compute the addition without carrying with a ^ b, and we find where we need to add carry bits with (a & b) << 1. Now, we just need to add those two together. Well, we already have a function for adding numbers together; it's called getSum. So, we basically just write function getSum(a, b) return getSum(a ^ b, (a & b) << 1); , except we make sure to short-circuit if there is nothing to carry, saving us from infinite recursion.






                share|improve this answer

























                  0












                  0








                  0







                  ^ is XOR, a bitwise operation. On a single bit, the rules are 0 ^ 0 = 0, 0 ^ 1 = 1, 1 ^ 0 = 0, and 1 ^ 1 = 0, and you simply extend perform it on corresponding bits when dealing with multi-bit values. The name is short for "exclusive or", and comes from the fact that A ^ B is 1 if and only if either A or B is 1, not both. But, it's more interesting to talk about its other name, ⊕. ⊕ is + but slightly different. You'll notice that the rules for ⊕ are similar to the rules for addition: 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, and 1 + 1 = 10. ⊕ is +, except 1 ⊕ 1 = 0; that is, ⊕ is +, except without carrying. This holds for multiple bits: 011 + 001 = 100, because you carry a 1 out of the ones place into the twos place, and then carry a 1 again into the fours place. Then, 011 ⊕ 001 = 010, because you just don't carry.



                  Now, when doing real addition, when do you carry? In binary, the answer is very simple: you carry a 1 into the next place when there are two 1s in a given place. This is easily understood as a bitwise AND, &. 1 & 1 = 1, and 0 otherwise. For 011 + 001, addition without carrying gives 011 ⊕ 001 = 010, and we can tell we need to carry a 1 out of the ones place because 011 & 001 = 001. The shifting in (a & b) << 1 turns a number "where do I need to carry from?" into "where do I need to add carries?": (011 & 001) << 1 = 010; I need to add a carry bit in the twos place.



                  So, in getSum, we want to know a + b. We compute the addition without carrying with a ^ b, and we find where we need to add carry bits with (a & b) << 1. Now, we just need to add those two together. Well, we already have a function for adding numbers together; it's called getSum. So, we basically just write function getSum(a, b) return getSum(a ^ b, (a & b) << 1); , except we make sure to short-circuit if there is nothing to carry, saving us from infinite recursion.






                  share|improve this answer













                  ^ is XOR, a bitwise operation. On a single bit, the rules are 0 ^ 0 = 0, 0 ^ 1 = 1, 1 ^ 0 = 0, and 1 ^ 1 = 0, and you simply extend perform it on corresponding bits when dealing with multi-bit values. The name is short for "exclusive or", and comes from the fact that A ^ B is 1 if and only if either A or B is 1, not both. But, it's more interesting to talk about its other name, ⊕. ⊕ is + but slightly different. You'll notice that the rules for ⊕ are similar to the rules for addition: 0 + 0 = 0, 0 + 1 = 1, 1 + 0 = 1, and 1 + 1 = 10. ⊕ is +, except 1 ⊕ 1 = 0; that is, ⊕ is +, except without carrying. This holds for multiple bits: 011 + 001 = 100, because you carry a 1 out of the ones place into the twos place, and then carry a 1 again into the fours place. Then, 011 ⊕ 001 = 010, because you just don't carry.



                  Now, when doing real addition, when do you carry? In binary, the answer is very simple: you carry a 1 into the next place when there are two 1s in a given place. This is easily understood as a bitwise AND, &. 1 & 1 = 1, and 0 otherwise. For 011 + 001, addition without carrying gives 011 ⊕ 001 = 010, and we can tell we need to carry a 1 out of the ones place because 011 & 001 = 001. The shifting in (a & b) << 1 turns a number "where do I need to carry from?" into "where do I need to add carries?": (011 & 001) << 1 = 010; I need to add a carry bit in the twos place.



                  So, in getSum, we want to know a + b. We compute the addition without carrying with a ^ b, and we find where we need to add carry bits with (a & b) << 1. Now, we just need to add those two together. Well, we already have a function for adding numbers together; it's called getSum. So, we basically just write function getSum(a, b) return getSum(a ^ b, (a & b) << 1); , except we make sure to short-circuit if there is nothing to carry, saving us from infinite recursion.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered yesterday









                  HTNWHTNW

                  10.1k1832




                  10.1k1832



























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Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029