Why is `const int& k = i; ++i; ` possible? [duplicate]
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
c++ syntax reference const
asked 14 hours ago
user9078057user9078057
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.
add a comment |
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
c++ syntax reference const
asked 14 hours ago
user9078057user9078057
1268
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 geterror: 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
add a comment |
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
c++ syntax reference const
asked 14 hours ago
user9078057user9078057
1268
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
c++ syntax reference const
asked 14 hours ago
user9078057user9078057
1268
asked 14 hours ago
user9078057user9078057
1268
asked 14 hours ago
user9078057user9078057
1268
asked 14 hours ago
user9078057user9078057
1268
asked 14 hours ago
user9078057user9078057
1268
1268
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.
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.
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 geterror: 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
add a comment |
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 geterror: 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
add a comment |
1 Answer
1
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oldest
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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.
answered 14 hours ago
R SahuR Sahu
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? sinceconst int *would be a non-constant pointer to a constant int andint * constwould be a constant pointer to a non-constant int, this should beint & const, shouldn't it? I feel like this is inconsistent.
– Max
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered 14 hours ago
R SahuR Sahu
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? sinceconst int *would be a non-constant pointer to a constant int andint * constwould be a constant pointer to a non-constant int, this should beint & const, shouldn't it? I feel like this is inconsistent.
– Max
3 hours ago
add a comment |
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.
answered 14 hours ago
R SahuR Sahu
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? sinceconst int *would be a non-constant pointer to a constant int andint * constwould be a constant pointer to a non-constant int, this should beint & const, shouldn't it? I feel like this is inconsistent.
– Max
3 hours ago
add a comment |
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.
answered 14 hours ago
R SahuR Sahu
169k1294193
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.
answered 14 hours ago
R SahuR Sahu
169k1294193
edited 14 hours ago
answered 14 hours ago
R SahuR Sahu
169k1294193
answered 14 hours ago
R SahuR Sahu
169k1294193
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? sinceconst int *would be a non-constant pointer to a constant int andint * constwould be a constant pointer to a non-constant int, this should beint & const, shouldn't it? I feel like this is inconsistent.
– Max
3 hours ago
add a comment |
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? sinceconst int *would be a non-constant pointer to a constant int andint * constwould be a constant pointer to a non-constant int, this should beint & 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
add a comment |
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