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What is the result of assigning to std::vector::begin()?
The Next CEO of Stack OverflowWhat are the differences between a pointer variable and a reference variable in C++?What does the explicit keyword mean?Concatenating two std::vectorsHow to find out if an item is present in a std::vector?Why is “using namespace std” considered bad practice?What is the “-->” operator in C++?What is the easiest way to initialize a std::vector with hardcoded elements?What is The Rule of Three?What are the basic rules and idioms for operator overloading?Why are std::begin and std::end “not memory safe”?
I've a bit understanding about lvalue and rvalues. So as I know we cannot assign to an rvalue but non-const lvalue is ok.
#include <iostream>
#include <vector>
int main()
std::vector<int> v 1, 2, 3, 4, 5 ;
v.begin() = v.end() - 2;
std::cout << *v.begin() << std::endl; // 1
for (auto const& e : v)
std::cout << e << ", ";// 1, 2, 3, 4, 5,
std::cout << std::endl;
Why I can assign to begin()
but it does nothing on the elements?
c++ vector
New contributor
add a comment |
I've a bit understanding about lvalue and rvalues. So as I know we cannot assign to an rvalue but non-const lvalue is ok.
#include <iostream>
#include <vector>
int main()
std::vector<int> v 1, 2, 3, 4, 5 ;
v.begin() = v.end() - 2;
std::cout << *v.begin() << std::endl; // 1
for (auto const& e : v)
std::cout << e << ", ";// 1, 2, 3, 4, 5,
std::cout << std::endl;
Why I can assign to begin()
but it does nothing on the elements?
c++ vector
New contributor
Why would you expect it to change anything in the vector? All you modify is an iterator
– UnholySheep
2 hours ago
4
It compiles if you include the necessary header files - I've added them. Why people think that including these in their question is optional, I will never know.
– Neil Butterworth
2 hours ago
@RSahu: Here is the output from Ideone: ideone.com/19AVFF
– Syfu_H
2 hours ago
add a comment |
I've a bit understanding about lvalue and rvalues. So as I know we cannot assign to an rvalue but non-const lvalue is ok.
#include <iostream>
#include <vector>
int main()
std::vector<int> v 1, 2, 3, 4, 5 ;
v.begin() = v.end() - 2;
std::cout << *v.begin() << std::endl; // 1
for (auto const& e : v)
std::cout << e << ", ";// 1, 2, 3, 4, 5,
std::cout << std::endl;
Why I can assign to begin()
but it does nothing on the elements?
c++ vector
New contributor
I've a bit understanding about lvalue and rvalues. So as I know we cannot assign to an rvalue but non-const lvalue is ok.
#include <iostream>
#include <vector>
int main()
std::vector<int> v 1, 2, 3, 4, 5 ;
v.begin() = v.end() - 2;
std::cout << *v.begin() << std::endl; // 1
for (auto const& e : v)
std::cout << e << ", ";// 1, 2, 3, 4, 5,
std::cout << std::endl;
Why I can assign to begin()
but it does nothing on the elements?
c++ vector
c++ vector
New contributor
New contributor
edited 2 hours ago
Neil Butterworth
27.2k54681
27.2k54681
New contributor
asked 2 hours ago
Syfu_HSyfu_H
493
493
New contributor
New contributor
Why would you expect it to change anything in the vector? All you modify is an iterator
– UnholySheep
2 hours ago
4
It compiles if you include the necessary header files - I've added them. Why people think that including these in their question is optional, I will never know.
– Neil Butterworth
2 hours ago
@RSahu: Here is the output from Ideone: ideone.com/19AVFF
– Syfu_H
2 hours ago
add a comment |
Why would you expect it to change anything in the vector? All you modify is an iterator
– UnholySheep
2 hours ago
4
It compiles if you include the necessary header files - I've added them. Why people think that including these in their question is optional, I will never know.
– Neil Butterworth
2 hours ago
@RSahu: Here is the output from Ideone: ideone.com/19AVFF
– Syfu_H
2 hours ago
Why would you expect it to change anything in the vector? All you modify is an iterator
– UnholySheep
2 hours ago
Why would you expect it to change anything in the vector? All you modify is an iterator
– UnholySheep
2 hours ago
4
4
It compiles if you include the necessary header files - I've added them. Why people think that including these in their question is optional, I will never know.
– Neil Butterworth
2 hours ago
It compiles if you include the necessary header files - I've added them. Why people think that including these in their question is optional, I will never know.
– Neil Butterworth
2 hours ago
@RSahu: Here is the output from Ideone: ideone.com/19AVFF
– Syfu_H
2 hours ago
@RSahu: Here is the output from Ideone: ideone.com/19AVFF
– Syfu_H
2 hours ago
add a comment |
3 Answers
3
active
oldest
votes
v.begin()
is an iterator. Assigning to an iterator does not assign to the value the iterator points to. In addition, because v.begin()
is a prvalue, the object that you are assigning to gets destroyed at the end of the line, without affecting v
or any of the "permanent" memory it owns.
If v.begin()
were a raw pointer, the compiler would issue an error for trying to assign to an rvalue. However, the type std::vector<T>::iterator
is often a class type, and it seems that this is the case on your implementation (with the specific set of compiler flags you are using), so assigning to v.begin()
calls the operator=
of that class type. But that doesn't change the fact that the iterator object is a temporary object, and assigning to it has no further effects.
In order to assign to the element that v.begin()
points to, you need to dereference it: *v.begin() = *(v.end() - 2)
1
What, no digression into lvalue qualified member functions?
– Yakk - Adam Nevraumont
2 hours ago
add a comment |
What is the result of assigning to std::vector::begin()?
The result of the assignment expression (the result is discarded in your example) is the assigned value. In this case the result is same as v.end() - 2
.
Note that since begin()
returns a value, this assignment only modifies the returned temporary object. The assignment doesn't modify the vector in any way. Given that the temporary object is discarded, the assignment has no observable effects in practice.
Futhermore, this assignment may be ill-formed if the standard library has chosen to implement std::vector::iterator
as a pointer.
Why I can assign to
begin()
Because
- Iterators are assignable.
- Rvalues of class type can be assigned †.
- The iterator happens to be a class type.
but it does nothing on the elements?
Because assigning an iterator doesn't modify the element that the iterator points at. Instead, the assignment changes what object the iterator is pointing at.
So as I know we cannot assign to an rvalue
This is not correct in general. In particular, it is not true for class types †.
† Unless the assignment operator of the class has lvalue-ref-qualifier (and has no rvalue-ref-qualified overload), which is unconventional.
How is a ref qualifier on an assignment operator "unconventional"? I'm unaware of any decent coding convention that fails to recommend it.
– Yakk - Adam Nevraumont
2 hours ago
@Yakk-AdamNevraumont It's not conventional for the standard library as far as I know. Can you name a class from standard library that uses it? I don't recall any, but that doesn't of course mean that they don't exist.
– eerorika
2 hours ago
@eerorika While looking into another question, I've just foundstd::optional::value
that has ref qualifiers. I don't remember seeing them anywhere else in the standard library though.
– François Andrieux
1 hour ago
What happened to produce "cdagger typoan"?
– genpfault
21 mins ago
@genpfault :D I usually google for dagger typo so that I can copy paste †. I don't know how I managed to get it into the answer.
– eerorika
10 mins ago
add a comment |
Sometimes code explains things better than words. Your code is equivalent to this:
std::vector<int> v 1, 2, 3, 4, 5 ;
auto temp = v.begin();
temp = v.end() - 2;
std::cout << *v.begin() << std::endl; // 1
for (auto const& e : v)
std::cout << e << ", ";// 1, 2, 3, 4, 5,
std::cout << std::endl;
In other words, v.begin() = v.end() - 2
has no observable effect. It's just an assignment to a temporary.
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
v.begin()
is an iterator. Assigning to an iterator does not assign to the value the iterator points to. In addition, because v.begin()
is a prvalue, the object that you are assigning to gets destroyed at the end of the line, without affecting v
or any of the "permanent" memory it owns.
If v.begin()
were a raw pointer, the compiler would issue an error for trying to assign to an rvalue. However, the type std::vector<T>::iterator
is often a class type, and it seems that this is the case on your implementation (with the specific set of compiler flags you are using), so assigning to v.begin()
calls the operator=
of that class type. But that doesn't change the fact that the iterator object is a temporary object, and assigning to it has no further effects.
In order to assign to the element that v.begin()
points to, you need to dereference it: *v.begin() = *(v.end() - 2)
1
What, no digression into lvalue qualified member functions?
– Yakk - Adam Nevraumont
2 hours ago
add a comment |
v.begin()
is an iterator. Assigning to an iterator does not assign to the value the iterator points to. In addition, because v.begin()
is a prvalue, the object that you are assigning to gets destroyed at the end of the line, without affecting v
or any of the "permanent" memory it owns.
If v.begin()
were a raw pointer, the compiler would issue an error for trying to assign to an rvalue. However, the type std::vector<T>::iterator
is often a class type, and it seems that this is the case on your implementation (with the specific set of compiler flags you are using), so assigning to v.begin()
calls the operator=
of that class type. But that doesn't change the fact that the iterator object is a temporary object, and assigning to it has no further effects.
In order to assign to the element that v.begin()
points to, you need to dereference it: *v.begin() = *(v.end() - 2)
1
What, no digression into lvalue qualified member functions?
– Yakk - Adam Nevraumont
2 hours ago
add a comment |
v.begin()
is an iterator. Assigning to an iterator does not assign to the value the iterator points to. In addition, because v.begin()
is a prvalue, the object that you are assigning to gets destroyed at the end of the line, without affecting v
or any of the "permanent" memory it owns.
If v.begin()
were a raw pointer, the compiler would issue an error for trying to assign to an rvalue. However, the type std::vector<T>::iterator
is often a class type, and it seems that this is the case on your implementation (with the specific set of compiler flags you are using), so assigning to v.begin()
calls the operator=
of that class type. But that doesn't change the fact that the iterator object is a temporary object, and assigning to it has no further effects.
In order to assign to the element that v.begin()
points to, you need to dereference it: *v.begin() = *(v.end() - 2)
v.begin()
is an iterator. Assigning to an iterator does not assign to the value the iterator points to. In addition, because v.begin()
is a prvalue, the object that you are assigning to gets destroyed at the end of the line, without affecting v
or any of the "permanent" memory it owns.
If v.begin()
were a raw pointer, the compiler would issue an error for trying to assign to an rvalue. However, the type std::vector<T>::iterator
is often a class type, and it seems that this is the case on your implementation (with the specific set of compiler flags you are using), so assigning to v.begin()
calls the operator=
of that class type. But that doesn't change the fact that the iterator object is a temporary object, and assigning to it has no further effects.
In order to assign to the element that v.begin()
points to, you need to dereference it: *v.begin() = *(v.end() - 2)
answered 2 hours ago
BrianBrian
66.2k798189
66.2k798189
1
What, no digression into lvalue qualified member functions?
– Yakk - Adam Nevraumont
2 hours ago
add a comment |
1
What, no digression into lvalue qualified member functions?
– Yakk - Adam Nevraumont
2 hours ago
1
1
What, no digression into lvalue qualified member functions?
– Yakk - Adam Nevraumont
2 hours ago
What, no digression into lvalue qualified member functions?
– Yakk - Adam Nevraumont
2 hours ago
add a comment |
What is the result of assigning to std::vector::begin()?
The result of the assignment expression (the result is discarded in your example) is the assigned value. In this case the result is same as v.end() - 2
.
Note that since begin()
returns a value, this assignment only modifies the returned temporary object. The assignment doesn't modify the vector in any way. Given that the temporary object is discarded, the assignment has no observable effects in practice.
Futhermore, this assignment may be ill-formed if the standard library has chosen to implement std::vector::iterator
as a pointer.
Why I can assign to
begin()
Because
- Iterators are assignable.
- Rvalues of class type can be assigned †.
- The iterator happens to be a class type.
but it does nothing on the elements?
Because assigning an iterator doesn't modify the element that the iterator points at. Instead, the assignment changes what object the iterator is pointing at.
So as I know we cannot assign to an rvalue
This is not correct in general. In particular, it is not true for class types †.
† Unless the assignment operator of the class has lvalue-ref-qualifier (and has no rvalue-ref-qualified overload), which is unconventional.
How is a ref qualifier on an assignment operator "unconventional"? I'm unaware of any decent coding convention that fails to recommend it.
– Yakk - Adam Nevraumont
2 hours ago
@Yakk-AdamNevraumont It's not conventional for the standard library as far as I know. Can you name a class from standard library that uses it? I don't recall any, but that doesn't of course mean that they don't exist.
– eerorika
2 hours ago
@eerorika While looking into another question, I've just foundstd::optional::value
that has ref qualifiers. I don't remember seeing them anywhere else in the standard library though.
– François Andrieux
1 hour ago
What happened to produce "cdagger typoan"?
– genpfault
21 mins ago
@genpfault :D I usually google for dagger typo so that I can copy paste †. I don't know how I managed to get it into the answer.
– eerorika
10 mins ago
add a comment |
What is the result of assigning to std::vector::begin()?
The result of the assignment expression (the result is discarded in your example) is the assigned value. In this case the result is same as v.end() - 2
.
Note that since begin()
returns a value, this assignment only modifies the returned temporary object. The assignment doesn't modify the vector in any way. Given that the temporary object is discarded, the assignment has no observable effects in practice.
Futhermore, this assignment may be ill-formed if the standard library has chosen to implement std::vector::iterator
as a pointer.
Why I can assign to
begin()
Because
- Iterators are assignable.
- Rvalues of class type can be assigned †.
- The iterator happens to be a class type.
but it does nothing on the elements?
Because assigning an iterator doesn't modify the element that the iterator points at. Instead, the assignment changes what object the iterator is pointing at.
So as I know we cannot assign to an rvalue
This is not correct in general. In particular, it is not true for class types †.
† Unless the assignment operator of the class has lvalue-ref-qualifier (and has no rvalue-ref-qualified overload), which is unconventional.
How is a ref qualifier on an assignment operator "unconventional"? I'm unaware of any decent coding convention that fails to recommend it.
– Yakk - Adam Nevraumont
2 hours ago
@Yakk-AdamNevraumont It's not conventional for the standard library as far as I know. Can you name a class from standard library that uses it? I don't recall any, but that doesn't of course mean that they don't exist.
– eerorika
2 hours ago
@eerorika While looking into another question, I've just foundstd::optional::value
that has ref qualifiers. I don't remember seeing them anywhere else in the standard library though.
– François Andrieux
1 hour ago
What happened to produce "cdagger typoan"?
– genpfault
21 mins ago
@genpfault :D I usually google for dagger typo so that I can copy paste †. I don't know how I managed to get it into the answer.
– eerorika
10 mins ago
add a comment |
What is the result of assigning to std::vector::begin()?
The result of the assignment expression (the result is discarded in your example) is the assigned value. In this case the result is same as v.end() - 2
.
Note that since begin()
returns a value, this assignment only modifies the returned temporary object. The assignment doesn't modify the vector in any way. Given that the temporary object is discarded, the assignment has no observable effects in practice.
Futhermore, this assignment may be ill-formed if the standard library has chosen to implement std::vector::iterator
as a pointer.
Why I can assign to
begin()
Because
- Iterators are assignable.
- Rvalues of class type can be assigned †.
- The iterator happens to be a class type.
but it does nothing on the elements?
Because assigning an iterator doesn't modify the element that the iterator points at. Instead, the assignment changes what object the iterator is pointing at.
So as I know we cannot assign to an rvalue
This is not correct in general. In particular, it is not true for class types †.
† Unless the assignment operator of the class has lvalue-ref-qualifier (and has no rvalue-ref-qualified overload), which is unconventional.
What is the result of assigning to std::vector::begin()?
The result of the assignment expression (the result is discarded in your example) is the assigned value. In this case the result is same as v.end() - 2
.
Note that since begin()
returns a value, this assignment only modifies the returned temporary object. The assignment doesn't modify the vector in any way. Given that the temporary object is discarded, the assignment has no observable effects in practice.
Futhermore, this assignment may be ill-formed if the standard library has chosen to implement std::vector::iterator
as a pointer.
Why I can assign to
begin()
Because
- Iterators are assignable.
- Rvalues of class type can be assigned †.
- The iterator happens to be a class type.
but it does nothing on the elements?
Because assigning an iterator doesn't modify the element that the iterator points at. Instead, the assignment changes what object the iterator is pointing at.
So as I know we cannot assign to an rvalue
This is not correct in general. In particular, it is not true for class types †.
† Unless the assignment operator of the class has lvalue-ref-qualifier (and has no rvalue-ref-qualified overload), which is unconventional.
edited 10 mins ago
answered 2 hours ago
eerorikaeerorika
88.2k663134
88.2k663134
How is a ref qualifier on an assignment operator "unconventional"? I'm unaware of any decent coding convention that fails to recommend it.
– Yakk - Adam Nevraumont
2 hours ago
@Yakk-AdamNevraumont It's not conventional for the standard library as far as I know. Can you name a class from standard library that uses it? I don't recall any, but that doesn't of course mean that they don't exist.
– eerorika
2 hours ago
@eerorika While looking into another question, I've just foundstd::optional::value
that has ref qualifiers. I don't remember seeing them anywhere else in the standard library though.
– François Andrieux
1 hour ago
What happened to produce "cdagger typoan"?
– genpfault
21 mins ago
@genpfault :D I usually google for dagger typo so that I can copy paste †. I don't know how I managed to get it into the answer.
– eerorika
10 mins ago
add a comment |
How is a ref qualifier on an assignment operator "unconventional"? I'm unaware of any decent coding convention that fails to recommend it.
– Yakk - Adam Nevraumont
2 hours ago
@Yakk-AdamNevraumont It's not conventional for the standard library as far as I know. Can you name a class from standard library that uses it? I don't recall any, but that doesn't of course mean that they don't exist.
– eerorika
2 hours ago
@eerorika While looking into another question, I've just foundstd::optional::value
that has ref qualifiers. I don't remember seeing them anywhere else in the standard library though.
– François Andrieux
1 hour ago
What happened to produce "cdagger typoan"?
– genpfault
21 mins ago
@genpfault :D I usually google for dagger typo so that I can copy paste †. I don't know how I managed to get it into the answer.
– eerorika
10 mins ago
How is a ref qualifier on an assignment operator "unconventional"? I'm unaware of any decent coding convention that fails to recommend it.
– Yakk - Adam Nevraumont
2 hours ago
How is a ref qualifier on an assignment operator "unconventional"? I'm unaware of any decent coding convention that fails to recommend it.
– Yakk - Adam Nevraumont
2 hours ago
@Yakk-AdamNevraumont It's not conventional for the standard library as far as I know. Can you name a class from standard library that uses it? I don't recall any, but that doesn't of course mean that they don't exist.
– eerorika
2 hours ago
@Yakk-AdamNevraumont It's not conventional for the standard library as far as I know. Can you name a class from standard library that uses it? I don't recall any, but that doesn't of course mean that they don't exist.
– eerorika
2 hours ago
@eerorika While looking into another question, I've just found
std::optional::value
that has ref qualifiers. I don't remember seeing them anywhere else in the standard library though.– François Andrieux
1 hour ago
@eerorika While looking into another question, I've just found
std::optional::value
that has ref qualifiers. I don't remember seeing them anywhere else in the standard library though.– François Andrieux
1 hour ago
What happened to produce "cdagger typoan"?
– genpfault
21 mins ago
What happened to produce "cdagger typoan"?
– genpfault
21 mins ago
@genpfault :D I usually google for dagger typo so that I can copy paste †. I don't know how I managed to get it into the answer.
– eerorika
10 mins ago
@genpfault :D I usually google for dagger typo so that I can copy paste †. I don't know how I managed to get it into the answer.
– eerorika
10 mins ago
add a comment |
Sometimes code explains things better than words. Your code is equivalent to this:
std::vector<int> v 1, 2, 3, 4, 5 ;
auto temp = v.begin();
temp = v.end() - 2;
std::cout << *v.begin() << std::endl; // 1
for (auto const& e : v)
std::cout << e << ", ";// 1, 2, 3, 4, 5,
std::cout << std::endl;
In other words, v.begin() = v.end() - 2
has no observable effect. It's just an assignment to a temporary.
add a comment |
Sometimes code explains things better than words. Your code is equivalent to this:
std::vector<int> v 1, 2, 3, 4, 5 ;
auto temp = v.begin();
temp = v.end() - 2;
std::cout << *v.begin() << std::endl; // 1
for (auto const& e : v)
std::cout << e << ", ";// 1, 2, 3, 4, 5,
std::cout << std::endl;
In other words, v.begin() = v.end() - 2
has no observable effect. It's just an assignment to a temporary.
add a comment |
Sometimes code explains things better than words. Your code is equivalent to this:
std::vector<int> v 1, 2, 3, 4, 5 ;
auto temp = v.begin();
temp = v.end() - 2;
std::cout << *v.begin() << std::endl; // 1
for (auto const& e : v)
std::cout << e << ", ";// 1, 2, 3, 4, 5,
std::cout << std::endl;
In other words, v.begin() = v.end() - 2
has no observable effect. It's just an assignment to a temporary.
Sometimes code explains things better than words. Your code is equivalent to this:
std::vector<int> v 1, 2, 3, 4, 5 ;
auto temp = v.begin();
temp = v.end() - 2;
std::cout << *v.begin() << std::endl; // 1
for (auto const& e : v)
std::cout << e << ", ";// 1, 2, 3, 4, 5,
std::cout << std::endl;
In other words, v.begin() = v.end() - 2
has no observable effect. It's just an assignment to a temporary.
answered 1 hour ago
Nikos C.Nikos C.
33.8k53967
33.8k53967
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Why would you expect it to change anything in the vector? All you modify is an iterator
– UnholySheep
2 hours ago
4
It compiles if you include the necessary header files - I've added them. Why people think that including these in their question is optional, I will never know.
– Neil Butterworth
2 hours ago
@RSahu: Here is the output from Ideone: ideone.com/19AVFF
– Syfu_H
2 hours ago