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 Mar 26 at 22:27
user9078057user9078057
1369
marked as duplicate by Cody Gray♦ c++
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Mar 27 at 7:13
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 Mar 26 at 22:27
user9078057user9078057
1369
marked as duplicate by Cody Gray♦ c++
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Mar 27 at 7:13
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
Mar 27 at 6:37
@danielspaniol I geterror: cannot assign to variable 'k' with const-qualified type 'const int &'
– Thomas Sablik
Mar 27 at 10:38
@ThomasSablik That is a semantical error. The syntax is correct.
– danielspaniol
Mar 27 at 10:44
I am setting the memory space of the newly created k. No. K is a reference, this means is equivalent to an alias (an alias is literally another another name for a variable). Sokdoes not need its own variable it is simply another name fori(though it is const qualified). You can verify that it does not have its own memory because you can not take the address ofk, if you try you will get the address ofi.
– Martin York
Mar 28 at 17:31
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 Mar 26 at 22:27
user9078057user9078057
1369
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 Mar 26 at 22:27
user9078057user9078057
1369
asked Mar 26 at 22:27
user9078057user9078057
1369
asked Mar 26 at 22:27
user9078057user9078057
1369
asked Mar 26 at 22:27
user9078057user9078057
1369
asked Mar 26 at 22:27
user9078057user9078057
1369
1369
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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Mar 27 at 7:13
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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Mar 27 at 7:13
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
Mar 27 at 6:37
@danielspaniol I geterror: cannot assign to variable 'k' with const-qualified type 'const int &'
– Thomas Sablik
Mar 27 at 10:38
@ThomasSablik That is a semantical error. The syntax is correct.
– danielspaniol
Mar 27 at 10:44
I am setting the memory space of the newly created k. No. K is a reference, this means is equivalent to an alias (an alias is literally another another name for a variable). Sokdoes not need its own variable it is simply another name fori(though it is const qualified). You can verify that it does not have its own memory because you can not take the address ofk, if you try you will get the address ofi.
– Martin York
Mar 28 at 17:31
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
Mar 27 at 6:37
@danielspaniol I geterror: cannot assign to variable 'k' with const-qualified type 'const int &'
– Thomas Sablik
Mar 27 at 10:38
@ThomasSablik That is a semantical error. The syntax is correct.
– danielspaniol
Mar 27 at 10:44
I am setting the memory space of the newly created k. No. K is a reference, this means is equivalent to an alias (an alias is literally another another name for a variable). Sokdoes not need its own variable it is simply another name fori(though it is const qualified). You can verify that it does not have its own memory because you can not take the address ofk, if you try you will get the address ofi.
– Martin York
Mar 28 at 17:31
1
1
Small nitpicking:
int f3(int i, int j) { const int& k=i; ++k; return k; } would be syntactically correct too– danielspaniol
Mar 27 at 6:37
Small nitpicking:
int f3(int i, int j) { const int& k=i; ++k; return k; } would be syntactically correct too– danielspaniol
Mar 27 at 6:37
@danielspaniol I get
error: cannot assign to variable 'k' with const-qualified type 'const int &'– Thomas Sablik
Mar 27 at 10:38
@danielspaniol I get
error: cannot assign to variable 'k' with const-qualified type 'const int &'– Thomas Sablik
Mar 27 at 10:38
@ThomasSablik That is a semantical error. The syntax is correct.
– danielspaniol
Mar 27 at 10:44
@ThomasSablik That is a semantical error. The syntax is correct.
– danielspaniol
Mar 27 at 10:44
I am setting the memory space of the newly created k. No. K is a reference, this means is equivalent to an alias (an alias is literally another another name for a variable). So k does not need its own variable it is simply another name for i (though it is const qualified). You can verify that it does not have its own memory because you can not take the address of k, if you try you will get the address of i.– Martin York
Mar 28 at 17:31
I am setting the memory space of the newly created k. No. K is a reference, this means is equivalent to an alias (an alias is literally another another name for a variable). So k does not need its own variable it is simply another name for i (though it is const qualified). You can verify that it does not have its own memory because you can not take the address of k, if you try you will get the address of i.– Martin York
Mar 28 at 17:31
add a comment |
1 Answer
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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 an analogy from a non-computer world.
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 Mar 26 at 22:31
R SahuR Sahu
170k1294193
7
Not so far-fetched. That's a good analogy.
– user4581301
Mar 26 at 22:44
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
Mar 26 at 22:53
@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
Mar 27 at 8:52
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 an analogy from a non-computer world.
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 Mar 26 at 22:31
R SahuR Sahu
170k1294193
7
Not so far-fetched. That's a good analogy.
– user4581301
Mar 26 at 22:44
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
Mar 26 at 22:53
@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
Mar 27 at 8:52
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 an analogy from a non-computer world.
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 Mar 26 at 22:31
R SahuR Sahu
170k1294193
7
Not so far-fetched. That's a good analogy.
– user4581301
Mar 26 at 22:44
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
Mar 26 at 22:53
@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
Mar 27 at 8:52
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 an analogy from a non-computer world.
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 Mar 26 at 22:31
R SahuR Sahu
170k1294193
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 an analogy from a non-computer world.
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 Mar 26 at 22:31
R SahuR Sahu
170k1294193
edited Mar 28 at 17:27
answered Mar 26 at 22:31
R SahuR Sahu
170k1294193
answered Mar 26 at 22:31
R SahuR Sahu
170k1294193
answered Mar 26 at 22:31
R SahuR Sahu
170k1294193
170k1294193
7
Not so far-fetched. That's a good analogy.
– user4581301
Mar 26 at 22:44
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
Mar 26 at 22:53
@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
Mar 27 at 8:52
add a comment |
7
Not so far-fetched. That's a good analogy.
– user4581301
Mar 26 at 22:44
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
Mar 26 at 22:53
@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
Mar 27 at 8:52
7
7
Not so far-fetched. That's a good analogy.
– user4581301
Mar 26 at 22:44
Not so far-fetched. That's a good analogy.
– user4581301
Mar 26 at 22:44
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
Mar 26 at 22:53
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
Mar 26 at 22:53
@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
Mar 27 at 8:52
@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
Mar 27 at 8:52
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
Mar 27 at 6:37
@danielspaniol I get
error: cannot assign to variable 'k' with const-qualified type 'const int &'– Thomas Sablik
Mar 27 at 10:38
@ThomasSablik That is a semantical error. The syntax is correct.
– danielspaniol
Mar 27 at 10:44
I am setting the memory space of the newly created k. No. K is a reference, this means is equivalent to an alias (an alias is literally another another name for a variable). Sokdoes not need its own variable it is simply another name fori(though it is const qualified). You can verify that it does not have its own memory because you can not take the address ofk, if you try you will get the address ofi.– Martin York
Mar 28 at 17:31