Why is `const int& k = i; ++i; ` possible? [duplicate]












9
















This question already has an answer here:




  • Why can const int& bind to an int?

    8 answers



  • A const & refers to a nonvolatile variable. The variable changes. Does the change invalidate the const &?

    4 answers




I am supposed to determine whether this function is syntactically correct:



int f3(int i, int j) { const int& k=i; ++i; return k; }



I have tested it out and it compiles with my main function.



I do not understand why this is so.



Surely by calling the function f3 I create copies of the variables iand j in a new memory space and setting const int& k=i I am setting the memory space of the newly created k to the exact the same space of the memory space of the copied i, therefore any change, i.e. the increment ++iwill result in ++k which is not possible given that it was set const



Any help is greatly appreciated










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5 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1





    Small nitpicking: int f3(int i, int j) { const int& k=i; ++k; return k; } would be syntactically correct too

    – danielspaniol
    5 hours ago











  • @danielspaniol I get error: cannot assign to variable 'k' with const-qualified type 'const int &'

    – Thomas Sablik
    1 hour ago











  • @ThomasSablik That is a semantical error. The syntax is correct.

    – danielspaniol
    1 hour ago
















9
















This question already has an answer here:




  • Why can const int& bind to an int?

    8 answers



  • A const & refers to a nonvolatile variable. The variable changes. Does the change invalidate the const &?

    4 answers




I am supposed to determine whether this function is syntactically correct:



int f3(int i, int j) { const int& k=i; ++i; return k; }



I have tested it out and it compiles with my main function.



I do not understand why this is so.



Surely by calling the function f3 I create copies of the variables iand j in a new memory space and setting const int& k=i I am setting the memory space of the newly created k to the exact the same space of the memory space of the copied i, therefore any change, i.e. the increment ++iwill result in ++k which is not possible given that it was set const



Any help is greatly appreciated










share|improve this question













marked as duplicate by Cody Gray c++
Users with the  c++ badge can single-handedly close c++ questions as duplicates and reopen them as needed.

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5 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • 1





    Small nitpicking: int f3(int i, int j) { const int& k=i; ++k; return k; } would be syntactically correct too

    – danielspaniol
    5 hours ago











  • @danielspaniol I get error: cannot assign to variable 'k' with const-qualified type 'const int &'

    – Thomas Sablik
    1 hour ago











  • @ThomasSablik That is a semantical error. The syntax is correct.

    – danielspaniol
    1 hour ago














9












9








9


1







This question already has an answer here:




  • Why can const int& bind to an int?

    8 answers



  • A const & refers to a nonvolatile variable. The variable changes. Does the change invalidate the const &?

    4 answers




I am supposed to determine whether this function is syntactically correct:



int f3(int i, int j) { const int& k=i; ++i; return k; }



I have tested it out and it compiles with my main function.



I do not understand why this is so.



Surely by calling the function f3 I create copies of the variables iand j in a new memory space and setting const int& k=i I am setting the memory space of the newly created k to the exact the same space of the memory space of the copied i, therefore any change, i.e. the increment ++iwill result in ++k which is not possible given that it was set const



Any help is greatly appreciated










share|improve this question















This question already has an answer here:




  • Why can const int& bind to an int?

    8 answers



  • A const & refers to a nonvolatile variable. The variable changes. Does the change invalidate the const &?

    4 answers




I am supposed to determine whether this function is syntactically correct:



int f3(int i, int j) { const int& k=i; ++i; return k; }



I have tested it out and it compiles with my main function.



I do not understand why this is so.



Surely by calling the function f3 I create copies of the variables iand j in a new memory space and setting const int& k=i I am setting the memory space of the newly created k to the exact the same space of the memory space of the copied i, therefore any change, i.e. the increment ++iwill result in ++k which is not possible given that it was set const



Any help is greatly appreciated





This question already has an answer here:




  • Why can const int& bind to an int?

    8 answers



  • A const & refers to a nonvolatile variable. The variable changes. Does the change invalidate the const &?

    4 answers








c++ syntax reference const






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 14 hours ago









user9078057user9078057

1268




1268




marked as duplicate by Cody Gray c++
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5 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









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5 hours ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 1





    Small nitpicking: int f3(int i, int j) { const int& k=i; ++k; return k; } would be syntactically correct too

    – danielspaniol
    5 hours ago











  • @danielspaniol I get error: cannot assign to variable 'k' with const-qualified type 'const int &'

    – Thomas Sablik
    1 hour ago











  • @ThomasSablik That is a semantical error. The syntax is correct.

    – danielspaniol
    1 hour ago














  • 1





    Small nitpicking: int f3(int i, int j) { const int& k=i; ++k; return k; } would be syntactically correct too

    – danielspaniol
    5 hours ago











  • @danielspaniol I get error: cannot assign to variable 'k' with const-qualified type 'const int &'

    – Thomas Sablik
    1 hour ago











  • @ThomasSablik That is a semantical error. The syntax is correct.

    – danielspaniol
    1 hour ago








1




1





Small nitpicking: int f3(int i, int j) { const int& k=i; ++k; return k; } would be syntactically correct too

– danielspaniol
5 hours ago





Small nitpicking: int f3(int i, int j) { const int& k=i; ++k; return k; } would be syntactically correct too

– danielspaniol
5 hours ago













@danielspaniol I get error: cannot assign to variable 'k' with const-qualified type 'const int &'

– Thomas Sablik
1 hour ago





@danielspaniol I get error: cannot assign to variable 'k' with const-qualified type 'const int &'

– Thomas Sablik
1 hour ago













@ThomasSablik That is a semantical error. The syntax is correct.

– danielspaniol
1 hour ago





@ThomasSablik That is a semantical error. The syntax is correct.

– danielspaniol
1 hour ago












1 Answer
1






active

oldest

votes


















20















the increment ++iwill result in ++k which is not possible given that it was set const




That's a misunderstanding.



You may not change the value of the object through k but it can still be changed through other means. In other words, ++k is not allowed but ++i is still allowed, which will indirectly modify the value of k.



Here's a far-fetched analogy.



You may look through the window of a store and see what's inside but you won't be able to change what's inside the store. However, an employee, who is inside the store, can change the contents of the store. You will see that change from
outside. You have const access or view access to the store while the employee has non-const access or change access to the store.






share|improve this answer





















  • 6





    Not so far-fetched. That's a good analogy.

    – user4581301
    13 hours ago











  • Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.

    – David Schwartz
    13 hours ago











  • @DavidSchwartz This is true, but now that I think of it, it doesn't suit the logic for pointers, does it? since const int * would be a non-constant pointer to a constant int and int * const would be a constant pointer to a non-constant int, this should be int & const, shouldn't it? I feel like this is inconsistent.

    – Max
    3 hours ago


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









20















the increment ++iwill result in ++k which is not possible given that it was set const




That's a misunderstanding.



You may not change the value of the object through k but it can still be changed through other means. In other words, ++k is not allowed but ++i is still allowed, which will indirectly modify the value of k.



Here's a far-fetched analogy.



You may look through the window of a store and see what's inside but you won't be able to change what's inside the store. However, an employee, who is inside the store, can change the contents of the store. You will see that change from
outside. You have const access or view access to the store while the employee has non-const access or change access to the store.






share|improve this answer





















  • 6





    Not so far-fetched. That's a good analogy.

    – user4581301
    13 hours ago











  • Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.

    – David Schwartz
    13 hours ago











  • @DavidSchwartz This is true, but now that I think of it, it doesn't suit the logic for pointers, does it? since const int * would be a non-constant pointer to a constant int and int * const would be a constant pointer to a non-constant int, this should be int & const, shouldn't it? I feel like this is inconsistent.

    – Max
    3 hours ago
















20















the increment ++iwill result in ++k which is not possible given that it was set const




That's a misunderstanding.



You may not change the value of the object through k but it can still be changed through other means. In other words, ++k is not allowed but ++i is still allowed, which will indirectly modify the value of k.



Here's a far-fetched analogy.



You may look through the window of a store and see what's inside but you won't be able to change what's inside the store. However, an employee, who is inside the store, can change the contents of the store. You will see that change from
outside. You have const access or view access to the store while the employee has non-const access or change access to the store.






share|improve this answer





















  • 6





    Not so far-fetched. That's a good analogy.

    – user4581301
    13 hours ago











  • Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.

    – David Schwartz
    13 hours ago











  • @DavidSchwartz This is true, but now that I think of it, it doesn't suit the logic for pointers, does it? since const int * would be a non-constant pointer to a constant int and int * const would be a constant pointer to a non-constant int, this should be int & const, shouldn't it? I feel like this is inconsistent.

    – Max
    3 hours ago














20












20








20








the increment ++iwill result in ++k which is not possible given that it was set const




That's a misunderstanding.



You may not change the value of the object through k but it can still be changed through other means. In other words, ++k is not allowed but ++i is still allowed, which will indirectly modify the value of k.



Here's a far-fetched analogy.



You may look through the window of a store and see what's inside but you won't be able to change what's inside the store. However, an employee, who is inside the store, can change the contents of the store. You will see that change from
outside. You have const access or view access to the store while the employee has non-const access or change access to the store.






share|improve this answer
















the increment ++iwill result in ++k which is not possible given that it was set const




That's a misunderstanding.



You may not change the value of the object through k but it can still be changed through other means. In other words, ++k is not allowed but ++i is still allowed, which will indirectly modify the value of k.



Here's a far-fetched analogy.



You may look through the window of a store and see what's inside but you won't be able to change what's inside the store. However, an employee, who is inside the store, can change the contents of the store. You will see that change from
outside. You have const access or view access to the store while the employee has non-const access or change access to the store.







share|improve this answer














share|improve this answer



share|improve this answer








edited 14 hours ago

























answered 14 hours ago









R SahuR Sahu

169k1294193




169k1294193








  • 6





    Not so far-fetched. That's a good analogy.

    – user4581301
    13 hours ago











  • Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.

    – David Schwartz
    13 hours ago











  • @DavidSchwartz This is true, but now that I think of it, it doesn't suit the logic for pointers, does it? since const int * would be a non-constant pointer to a constant int and int * const would be a constant pointer to a non-constant int, this should be int & const, shouldn't it? I feel like this is inconsistent.

    – Max
    3 hours ago














  • 6





    Not so far-fetched. That's a good analogy.

    – user4581301
    13 hours ago











  • Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.

    – David Schwartz
    13 hours ago











  • @DavidSchwartz This is true, but now that I think of it, it doesn't suit the logic for pointers, does it? since const int * would be a non-constant pointer to a constant int and int * const would be a constant pointer to a non-constant int, this should be int & const, shouldn't it? I feel like this is inconsistent.

    – Max
    3 hours ago








6




6





Not so far-fetched. That's a good analogy.

– user4581301
13 hours ago





Not so far-fetched. That's a good analogy.

– user4581301
13 hours ago













Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.

– David Schwartz
13 hours ago





Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.

– David Schwartz
13 hours ago













@DavidSchwartz This is true, but now that I think of it, it doesn't suit the logic for pointers, does it? since const int * would be a non-constant pointer to a constant int and int * const would be a constant pointer to a non-constant int, this should be int & const, shouldn't it? I feel like this is inconsistent.

– Max
3 hours ago





@DavidSchwartz This is true, but now that I think of it, it doesn't suit the logic for pointers, does it? since const int * would be a non-constant pointer to a constant int and int * const would be a constant pointer to a non-constant int, this should be int & const, shouldn't it? I feel like this is inconsistent.

– Max
3 hours ago





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