C++ const map element access
I tried to use the operator access the element in a const C++ map, but this method failed. I also tried to use "at()" to do the same thing. It worked this time. However, I could not find any reference about using "at()" to access element in a const C++ map. Is "at()" a newly added function in C++ map? Where can I find more info about this? Thank you very much!
An example could be the following:
#include <iostream>
#include <map>
using namespace std;
int main()
{
map<int, char> A;
A[1] = 'b';
A[3] = 'c';
const map<int, char> B = A;
cout << B.at(3) << endl; // it works
cout << B[3] << endl; // it does not work
}
For using "B[3]", it returned the following errors during compiling:
t01.cpp:14: error: passing ‘const
std::map,
std::allocator > >’ as ‘this’ argument of ‘_Tp&
std::map<_Key, _Tp, _Compare,
_Alloc>::operator(const _Key&) [with _Key = int, _Tp = char, _Compare = std::less, _Alloc =
std::allocator >]’ discards qualifiers
The compiler used is g++ 4.2.1
c++ stl map const
asked Feb 27 '11 at 17:17
icephereicephere
465145
add a comment |
I tried to use the operator access the element in a const C++ map, but this method failed. I also tried to use "at()" to do the same thing. It worked this time. However, I could not find any reference about using "at()" to access element in a const C++ map. Is "at()" a newly added function in C++ map? Where can I find more info about this? Thank you very much!
An example could be the following:
#include <iostream>
#include <map>
using namespace std;
int main()
{
map<int, char> A;
A[1] = 'b';
A[3] = 'c';
const map<int, char> B = A;
cout << B.at(3) << endl; // it works
cout << B[3] << endl; // it does not work
}
For using "B[3]", it returned the following errors during compiling:
t01.cpp:14: error: passing ‘const
std::map,
std::allocator > >’ as ‘this’ argument of ‘_Tp&
std::map<_Key, _Tp, _Compare,
_Alloc>::operator(const _Key&) [with _Key = int, _Tp = char, _Compare = std::less, _Alloc =
std::allocator >]’ discards qualifiers
The compiler used is g++ 4.2.1
c++ stl map const
asked Feb 27 '11 at 17:17
icephereicephere
465145
add a comment |
I tried to use the operator access the element in a const C++ map, but this method failed. I also tried to use "at()" to do the same thing. It worked this time. However, I could not find any reference about using "at()" to access element in a const C++ map. Is "at()" a newly added function in C++ map? Where can I find more info about this? Thank you very much!
An example could be the following:
#include <iostream>
#include <map>
using namespace std;
int main()
{
map<int, char> A;
A[1] = 'b';
A[3] = 'c';
const map<int, char> B = A;
cout << B.at(3) << endl; // it works
cout << B[3] << endl; // it does not work
}
For using "B[3]", it returned the following errors during compiling:
t01.cpp:14: error: passing ‘const
std::map,
std::allocator > >’ as ‘this’ argument of ‘_Tp&
std::map<_Key, _Tp, _Compare,
_Alloc>::operator(const _Key&) [with _Key = int, _Tp = char, _Compare = std::less, _Alloc =
std::allocator >]’ discards qualifiers
The compiler used is g++ 4.2.1
c++ stl map const
asked Feb 27 '11 at 17:17
icephereicephere
465145
I tried to use the operator access the element in a const C++ map, but this method failed. I also tried to use "at()" to do the same thing. It worked this time. However, I could not find any reference about using "at()" to access element in a const C++ map. Is "at()" a newly added function in C++ map? Where can I find more info about this? Thank you very much!
An example could be the following:
#include <iostream>
#include <map>
using namespace std;
int main()
{
map<int, char> A;
A[1] = 'b';
A[3] = 'c';
const map<int, char> B = A;
cout << B.at(3) << endl; // it works
cout << B[3] << endl; // it does not work
}
For using "B[3]", it returned the following errors during compiling:
t01.cpp:14: error: passing ‘const
std::map,
std::allocator > >’ as ‘this’ argument of ‘_Tp&
std::map<_Key, _Tp, _Compare,
_Alloc>::operator(const _Key&) [with _Key = int, _Tp = char, _Compare = std::less, _Alloc =
std::allocator >]’ discards qualifiers
The compiler used is g++ 4.2.1
c++ stl map const
c++ stl map const
asked Feb 27 '11 at 17:17
icephereicephere
465145
asked Feb 27 '11 at 17:17
icephereicephere
465145
asked Feb 27 '11 at 17:17
icephereicephere
465145
asked Feb 27 '11 at 17:17
icephereicephere
465145
asked Feb 27 '11 at 17:17
icephereicephere
465145
465145
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
at() is a new method for std::map in C++11.
Rather than insert a new default constructed element as operator does if an element with the given key does not exist, it throws a std::out_of_range exception. (This is similar to the behaviour of at() for deque and vector.)
Because of this behaviour it makes sense for there to be a const overload of at(), unlike operator which always has the potential to change the map.
answered Feb 27 '11 at 17:29
CB BaileyCB Bailey
521k78559614
Is is possible to have "at" return a default value instead of throwing an exception?
– user1202136
Aug 30 '12 at 14:41
at()should be C++11 only
– Deqing
Aug 9 '13 at 10:54
I'm usingat()with in VS2013 on a project set to use VS2010 toolkit. I thought that meant I wasn't using C++11... But yet it compiles... ??
– thomthom
Dec 7 '13 at 22:27
1
I just need to comment that it doesn't make sense to omit the const operator, which could also throw an exception for an unmapped element instead of changing the map.
– Spencer
Mar 1 '18 at 15:37
@Spencer It would be surprising if the const and non-const overloads of operator had different effects. Normally, we expect that if some non-const objects or references in a program are made const, the program will continue to behave in the same way, as long as it compiles. Allowing only the non-const overload to throw exceptions can result in bugs that are not caught until runtime.
– Brian
Feb 25 at 17:55
|
show 7 more comments
If an element doesn’t exist in a map, the operator will add it – which obviously cannot work in a const map so C++ does not define a const version of the operator. This is a nice example of the compiler’s type checker preventing a potential runtime error.
In your case, you need to use find instead which will only return an (iterator to the) element if it exists, it will never modify the map. If an item doesn’t exist, it returns an iterator to the map’s end().
at doesn’t exist and shouldn’t even compile. Perhaps this is a “compiler extension” (= a bug new in C++0x).
answered Feb 27 '11 at 17:24
Konrad RudolphKonrad Rudolph
401k1017901038
Does the C++ standard forbid the implementation from defining additional non-standard member functions in library classes?
– Tim Martin
Feb 27 '11 at 17:31
@Tim I believe the interface is fixed, yes.
– Konrad Rudolph
Feb 27 '11 at 17:33
add a comment |
The -operator will create a new entry in the map if the given key does not exists. It may thus change the map.
See this link.
answered Feb 27 '11 at 17:28
vidstigevidstige
8,49054487
add a comment |
This comes as quite a surprise to me, but the STL map doesn't have a const index operator. That is, B[3] cannot be read-only. From the manual:
Since operator might insert a new element into the map, it can't possibly be a const member function.
I have no idea about at().
answered Feb 27 '11 at 17:27
BetaBeta
72.5k7110129
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
at() is a new method for std::map in C++11.
Rather than insert a new default constructed element as operator does if an element with the given key does not exist, it throws a std::out_of_range exception. (This is similar to the behaviour of at() for deque and vector.)
Because of this behaviour it makes sense for there to be a const overload of at(), unlike operator which always has the potential to change the map.
answered Feb 27 '11 at 17:29
CB BaileyCB Bailey
521k78559614
Is is possible to have "at" return a default value instead of throwing an exception?
– user1202136
Aug 30 '12 at 14:41
at()should be C++11 only
– Deqing
Aug 9 '13 at 10:54
I'm usingat()with in VS2013 on a project set to use VS2010 toolkit. I thought that meant I wasn't using C++11... But yet it compiles... ??
– thomthom
Dec 7 '13 at 22:27
1
I just need to comment that it doesn't make sense to omit the const operator, which could also throw an exception for an unmapped element instead of changing the map.
– Spencer
Mar 1 '18 at 15:37
@Spencer It would be surprising if the const and non-const overloads of operator had different effects. Normally, we expect that if some non-const objects or references in a program are made const, the program will continue to behave in the same way, as long as it compiles. Allowing only the non-const overload to throw exceptions can result in bugs that are not caught until runtime.
– Brian
Feb 25 at 17:55
|
show 7 more comments
at() is a new method for std::map in C++11.
Rather than insert a new default constructed element as operator does if an element with the given key does not exist, it throws a std::out_of_range exception. (This is similar to the behaviour of at() for deque and vector.)
Because of this behaviour it makes sense for there to be a const overload of at(), unlike operator which always has the potential to change the map.
answered Feb 27 '11 at 17:29
CB BaileyCB Bailey
521k78559614
Is is possible to have "at" return a default value instead of throwing an exception?
– user1202136
Aug 30 '12 at 14:41
at()should be C++11 only
– Deqing
Aug 9 '13 at 10:54
I'm usingat()with in VS2013 on a project set to use VS2010 toolkit. I thought that meant I wasn't using C++11... But yet it compiles... ??
– thomthom
Dec 7 '13 at 22:27
1
I just need to comment that it doesn't make sense to omit the const operator, which could also throw an exception for an unmapped element instead of changing the map.
– Spencer
Mar 1 '18 at 15:37
@Spencer It would be surprising if the const and non-const overloads of operator had different effects. Normally, we expect that if some non-const objects or references in a program are made const, the program will continue to behave in the same way, as long as it compiles. Allowing only the non-const overload to throw exceptions can result in bugs that are not caught until runtime.
– Brian
Feb 25 at 17:55
|
show 7 more comments
at() is a new method for std::map in C++11.
Rather than insert a new default constructed element as operator does if an element with the given key does not exist, it throws a std::out_of_range exception. (This is similar to the behaviour of at() for deque and vector.)
Because of this behaviour it makes sense for there to be a const overload of at(), unlike operator which always has the potential to change the map.
answered Feb 27 '11 at 17:29
CB BaileyCB Bailey
521k78559614
at() is a new method for std::map in C++11.
Rather than insert a new default constructed element as operator does if an element with the given key does not exist, it throws a std::out_of_range exception. (This is similar to the behaviour of at() for deque and vector.)
Because of this behaviour it makes sense for there to be a const overload of at(), unlike operator which always has the potential to change the map.
answered Feb 27 '11 at 17:29
CB BaileyCB Bailey
521k78559614
edited Aug 28 '13 at 14:38
user283145
answered Feb 27 '11 at 17:29
CB BaileyCB Bailey
521k78559614
answered Feb 27 '11 at 17:29
CB BaileyCB Bailey
521k78559614
answered Feb 27 '11 at 17:29
CB BaileyCB Bailey
521k78559614
521k78559614
Is is possible to have "at" return a default value instead of throwing an exception?
– user1202136
Aug 30 '12 at 14:41
at()should be C++11 only
– Deqing
Aug 9 '13 at 10:54
I'm usingat()with in VS2013 on a project set to use VS2010 toolkit. I thought that meant I wasn't using C++11... But yet it compiles... ??
– thomthom
Dec 7 '13 at 22:27
1
I just need to comment that it doesn't make sense to omit the const operator, which could also throw an exception for an unmapped element instead of changing the map.
– Spencer
Mar 1 '18 at 15:37
@Spencer It would be surprising if the const and non-const overloads of operator had different effects. Normally, we expect that if some non-const objects or references in a program are made const, the program will continue to behave in the same way, as long as it compiles. Allowing only the non-const overload to throw exceptions can result in bugs that are not caught until runtime.
– Brian
Feb 25 at 17:55
|
show 7 more comments
Is is possible to have "at" return a default value instead of throwing an exception?
– user1202136
Aug 30 '12 at 14:41
at()should be C++11 only
– Deqing
Aug 9 '13 at 10:54
I'm usingat()with in VS2013 on a project set to use VS2010 toolkit. I thought that meant I wasn't using C++11... But yet it compiles... ??
– thomthom
Dec 7 '13 at 22:27
1
I just need to comment that it doesn't make sense to omit the const operator, which could also throw an exception for an unmapped element instead of changing the map.
– Spencer
Mar 1 '18 at 15:37
@Spencer It would be surprising if the const and non-const overloads of operator had different effects. Normally, we expect that if some non-const objects or references in a program are made const, the program will continue to behave in the same way, as long as it compiles. Allowing only the non-const overload to throw exceptions can result in bugs that are not caught until runtime.
– Brian
Feb 25 at 17:55
Is is possible to have "at" return a default value instead of throwing an exception?
– user1202136
Aug 30 '12 at 14:41
Is is possible to have "at" return a default value instead of throwing an exception?
– user1202136
Aug 30 '12 at 14:41
at() should be C++11 only– Deqing
Aug 9 '13 at 10:54
at() should be C++11 only– Deqing
Aug 9 '13 at 10:54
I'm using
at() with in VS2013 on a project set to use VS2010 toolkit. I thought that meant I wasn't using C++11... But yet it compiles... ??– thomthom
Dec 7 '13 at 22:27
I'm using
at() with in VS2013 on a project set to use VS2010 toolkit. I thought that meant I wasn't using C++11... But yet it compiles... ??– thomthom
Dec 7 '13 at 22:27
1
1
I just need to comment that it doesn't make sense to omit the const operator, which could also throw an exception for an unmapped element instead of changing the map.
– Spencer
Mar 1 '18 at 15:37
I just need to comment that it doesn't make sense to omit the const operator, which could also throw an exception for an unmapped element instead of changing the map.
– Spencer
Mar 1 '18 at 15:37
@Spencer It would be surprising if the const and non-const overloads of operator had different effects. Normally, we expect that if some non-const objects or references in a program are made const, the program will continue to behave in the same way, as long as it compiles. Allowing only the non-const overload to throw exceptions can result in bugs that are not caught until runtime.
– Brian
Feb 25 at 17:55
@Spencer It would be surprising if the const and non-const overloads of operator had different effects. Normally, we expect that if some non-const objects or references in a program are made const, the program will continue to behave in the same way, as long as it compiles. Allowing only the non-const overload to throw exceptions can result in bugs that are not caught until runtime.
– Brian
Feb 25 at 17:55
|
show 7 more comments
If an element doesn’t exist in a map, the operator will add it – which obviously cannot work in a const map so C++ does not define a const version of the operator. This is a nice example of the compiler’s type checker preventing a potential runtime error.
In your case, you need to use find instead which will only return an (iterator to the) element if it exists, it will never modify the map. If an item doesn’t exist, it returns an iterator to the map’s end().
at doesn’t exist and shouldn’t even compile. Perhaps this is a “compiler extension” (= a bug new in C++0x).
answered Feb 27 '11 at 17:24
Konrad RudolphKonrad Rudolph
401k1017901038
Does the C++ standard forbid the implementation from defining additional non-standard member functions in library classes?
– Tim Martin
Feb 27 '11 at 17:31
@Tim I believe the interface is fixed, yes.
– Konrad Rudolph
Feb 27 '11 at 17:33
add a comment |
If an element doesn’t exist in a map, the operator will add it – which obviously cannot work in a const map so C++ does not define a const version of the operator. This is a nice example of the compiler’s type checker preventing a potential runtime error.
In your case, you need to use find instead which will only return an (iterator to the) element if it exists, it will never modify the map. If an item doesn’t exist, it returns an iterator to the map’s end().
at doesn’t exist and shouldn’t even compile. Perhaps this is a “compiler extension” (= a bug new in C++0x).
answered Feb 27 '11 at 17:24
Konrad RudolphKonrad Rudolph
401k1017901038
Does the C++ standard forbid the implementation from defining additional non-standard member functions in library classes?
– Tim Martin
Feb 27 '11 at 17:31
@Tim I believe the interface is fixed, yes.
– Konrad Rudolph
Feb 27 '11 at 17:33
add a comment |
If an element doesn’t exist in a map, the operator will add it – which obviously cannot work in a const map so C++ does not define a const version of the operator. This is a nice example of the compiler’s type checker preventing a potential runtime error.
In your case, you need to use find instead which will only return an (iterator to the) element if it exists, it will never modify the map. If an item doesn’t exist, it returns an iterator to the map’s end().
at doesn’t exist and shouldn’t even compile. Perhaps this is a “compiler extension” (= a bug new in C++0x).
answered Feb 27 '11 at 17:24
Konrad RudolphKonrad Rudolph
401k1017901038
If an element doesn’t exist in a map, the operator will add it – which obviously cannot work in a const map so C++ does not define a const version of the operator. This is a nice example of the compiler’s type checker preventing a potential runtime error.
In your case, you need to use find instead which will only return an (iterator to the) element if it exists, it will never modify the map. If an item doesn’t exist, it returns an iterator to the map’s end().
at doesn’t exist and shouldn’t even compile. Perhaps this is a “compiler extension” (= a bug new in C++0x).
answered Feb 27 '11 at 17:24
Konrad RudolphKonrad Rudolph
401k1017901038
edited Feb 27 '11 at 17:34
answered Feb 27 '11 at 17:24
Konrad RudolphKonrad Rudolph
401k1017901038
answered Feb 27 '11 at 17:24
Konrad RudolphKonrad Rudolph
401k1017901038
answered Feb 27 '11 at 17:24
Konrad RudolphKonrad Rudolph
401k1017901038
401k1017901038
Does the C++ standard forbid the implementation from defining additional non-standard member functions in library classes?
– Tim Martin
Feb 27 '11 at 17:31
@Tim I believe the interface is fixed, yes.
– Konrad Rudolph
Feb 27 '11 at 17:33
add a comment |
Does the C++ standard forbid the implementation from defining additional non-standard member functions in library classes?
– Tim Martin
Feb 27 '11 at 17:31
@Tim I believe the interface is fixed, yes.
– Konrad Rudolph
Feb 27 '11 at 17:33
Does the C++ standard forbid the implementation from defining additional non-standard member functions in library classes?
– Tim Martin
Feb 27 '11 at 17:31
Does the C++ standard forbid the implementation from defining additional non-standard member functions in library classes?
– Tim Martin
Feb 27 '11 at 17:31
@Tim I believe the interface is fixed, yes.
– Konrad Rudolph
Feb 27 '11 at 17:33
@Tim I believe the interface is fixed, yes.
– Konrad Rudolph
Feb 27 '11 at 17:33
add a comment |
The -operator will create a new entry in the map if the given key does not exists. It may thus change the map.
See this link.
answered Feb 27 '11 at 17:28
vidstigevidstige
8,49054487
add a comment |
The -operator will create a new entry in the map if the given key does not exists. It may thus change the map.
See this link.
answered Feb 27 '11 at 17:28
vidstigevidstige
8,49054487
add a comment |
The -operator will create a new entry in the map if the given key does not exists. It may thus change the map.
See this link.
answered Feb 27 '11 at 17:28
vidstigevidstige
8,49054487
The -operator will create a new entry in the map if the given key does not exists. It may thus change the map.
See this link.
answered Feb 27 '11 at 17:28
vidstigevidstige
8,49054487
answered Feb 27 '11 at 17:28
vidstigevidstige
8,49054487
answered Feb 27 '11 at 17:28
vidstigevidstige
8,49054487
answered Feb 27 '11 at 17:28
vidstigevidstige
8,49054487
8,49054487
add a comment |
add a comment |
This comes as quite a surprise to me, but the STL map doesn't have a const index operator. That is, B[3] cannot be read-only. From the manual:
Since operator might insert a new element into the map, it can't possibly be a const member function.
I have no idea about at().
answered Feb 27 '11 at 17:27
BetaBeta
72.5k7110129
add a comment |
This comes as quite a surprise to me, but the STL map doesn't have a const index operator. That is, B[3] cannot be read-only. From the manual:
Since operator might insert a new element into the map, it can't possibly be a const member function.
I have no idea about at().
answered Feb 27 '11 at 17:27
BetaBeta
72.5k7110129
add a comment |
This comes as quite a surprise to me, but the STL map doesn't have a const index operator. That is, B[3] cannot be read-only. From the manual:
Since operator might insert a new element into the map, it can't possibly be a const member function.
I have no idea about at().
answered Feb 27 '11 at 17:27
BetaBeta
72.5k7110129
This comes as quite a surprise to me, but the STL map doesn't have a const index operator. That is, B[3] cannot be read-only. From the manual:
Since operator might insert a new element into the map, it can't possibly be a const member function.
I have no idea about at().
answered Feb 27 '11 at 17:27
BetaBeta
72.5k7110129
answered Feb 27 '11 at 17:27
BetaBeta
72.5k7110129
answered Feb 27 '11 at 17:27
BetaBeta
72.5k7110129
answered Feb 27 '11 at 17:27
BetaBeta
72.5k7110129
72.5k7110129
add a comment |
add a comment |
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