Why is `const int& k = i; ++i; ` possible? [duplicate] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Data science time! April 2019 and salary with experience The Ask Question Wizard is Live!Why can const int& bind to an int?A const & refers to a nonvolatile variable. The variable changes. Does the change invalidate the const &?What is the difference between const and readonly?How to convert a std::string to const char* or char*?Why can templates only be implemented in the header file?Meaning of 'const' last in a function declaration of a class?What is the difference between const int*, const int * const, and int const *?Why is “using namespace std” considered bad practice?define() vs constWhy does ++[[]][+[]]+[+[]] return the string “10”?Why are elementwise additions much faster in separate loops than in a combined loop?Why is it faster to process a sorted array than an unsorted array?
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Why is `const int& k = i; ++i; ` possible? [duplicate]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Data science time! April 2019 and salary with experience
The Ask Question Wizard is Live!Why can const int& bind to an int?A const & refers to a nonvolatile variable. The variable changes. Does the change invalidate the const &?What is the difference between const and readonly?How to convert a std::string to const char* or char*?Why can templates only be implemented in the header file?Meaning of 'const' last in a function declaration of a class?What is the difference between const int*, const int * const, and int const *?Why is “using namespace std” considered bad practice?define() vs constWhy does ++[[]][+[]]+[+[]] return the string “10”?Why are elementwise additions much faster in separate loops than in a combined loop?Why is it faster to process a sorted array than an unsorted array?
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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
1469
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.
|
show 6 more comments
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
1469
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.
2
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
1
@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
@MartinYork If you putint &kin a structA, make it the first field, take address of the struct, and cast it fromA*toint**: you will have the address of theint &k.
– atomsymbol
Apr 3 at 22:18
|
show 6 more comments
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
1469
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
1469
asked Mar 26 at 22:27
user9078057user9078057
1469
asked Mar 26 at 22:27
user9078057user9078057
1469
asked Mar 26 at 22:27
user9078057user9078057
1469
asked Mar 26 at 22:27
user9078057user9078057
1469
1469
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.
2
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
1
@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
@MartinYork If you putint &kin a structA, make it the first field, take address of the struct, and cast it fromA*toint**: you will have the address of theint &k.
– atomsymbol
Apr 3 at 22:18
|
show 6 more comments
2
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
1
@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
@MartinYork If you putint &kin a structA, make it the first field, take address of the struct, and cast it fromA*toint**: you will have the address of theint &k.
– atomsymbol
Apr 3 at 22:18
2
2
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
1
1
@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
@MartinYork If you put
int &k in a struct A, make it the first field, take address of the struct, and cast it from A* to int**: you will have the address of the int &k.– atomsymbol
Apr 3 at 22:18
@MartinYork If you put
int &k in a struct A, make it the first field, take address of the struct, and cast it from A* to int**: you will have the address of the int &k.– atomsymbol
Apr 3 at 22:18
|
show 6 more comments
1 Answer
1
active
oldest
votes
the increment
++iwill result in++kwhich is not possible given that it was setconst
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.
edited Apr 3 at 7:22
El Profesor
11k32341
answered Mar 26 at 22:31
R SahuR Sahu
171k1298197
9
Not so far-fetched. That's a good analogy.
– user4581301
Mar 26 at 22:44
1
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++kwhich is not possible given that it was setconst
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.
edited Apr 3 at 7:22
El Profesor
11k32341
answered Mar 26 at 22:31
R SahuR Sahu
171k1298197
9
Not so far-fetched. That's a good analogy.
– user4581301
Mar 26 at 22:44
1
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++kwhich is not possible given that it was setconst
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.
edited Apr 3 at 7:22
El Profesor
11k32341
answered Mar 26 at 22:31
R SahuR Sahu
171k1298197
9
Not so far-fetched. That's a good analogy.
– user4581301
Mar 26 at 22:44
1
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++kwhich is not possible given that it was setconst
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.
edited Apr 3 at 7:22
El Profesor
11k32341
answered Mar 26 at 22:31
R SahuR Sahu
171k1298197
the increment
++iwill result in++kwhich is not possible given that it was setconst
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.
edited Apr 3 at 7:22
El Profesor
11k32341
answered Mar 26 at 22:31
R SahuR Sahu
171k1298197
edited Apr 3 at 7:22
El Profesor
11k32341
edited Apr 3 at 7:22
El Profesor
11k32341
edited Apr 3 at 7:22
El Profesor
11k32341
11k32341
answered Mar 26 at 22:31
R SahuR Sahu
171k1298197
answered Mar 26 at 22:31
R SahuR Sahu
171k1298197
answered Mar 26 at 22:31
R SahuR Sahu
171k1298197
171k1298197
9
Not so far-fetched. That's a good analogy.
– user4581301
Mar 26 at 22:44
1
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 |
9
Not so far-fetched. That's a good analogy.
– user4581301
Mar 26 at 22:44
1
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
9
9
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
1
1
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 |
2
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
1
@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
@MartinYork If you put
int &kin a structA, make it the first field, take address of the struct, and cast it fromA*toint**: you will have the address of theint &k.– atomsymbol
Apr 3 at 22:18