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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”?










5















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?










share|improve this question









New contributor




Syfu_H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • Why would you expect it to change anything in the vector? All you modify is an iterator

    – UnholySheep
    1 hour ago






  • 3





    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
    1 hour ago












  • @RSahu: Here is the output from Ideone: ideone.com/19AVFF

    – Syfu_H
    1 hour ago















5















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?










share|improve this question









New contributor




Syfu_H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Why would you expect it to change anything in the vector? All you modify is an iterator

    – UnholySheep
    1 hour ago






  • 3





    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
    1 hour ago












  • @RSahu: Here is the output from Ideone: ideone.com/19AVFF

    – Syfu_H
    1 hour ago













5












5








5


5






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?










share|improve this question









New contributor




Syfu_H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












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






share|improve this question









New contributor




Syfu_H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Syfu_H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 1 hour ago









Neil Butterworth

27.2k54681




27.2k54681






New contributor




Syfu_H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 1 hour ago









Syfu_HSyfu_H

443




443




New contributor




Syfu_H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Syfu_H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Syfu_H is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Why would you expect it to change anything in the vector? All you modify is an iterator

    – UnholySheep
    1 hour ago






  • 3





    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
    1 hour ago












  • @RSahu: Here is the output from Ideone: ideone.com/19AVFF

    – Syfu_H
    1 hour ago

















  • Why would you expect it to change anything in the vector? All you modify is an iterator

    – UnholySheep
    1 hour ago






  • 3





    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
    1 hour ago












  • @RSahu: Here is the output from Ideone: ideone.com/19AVFF

    – Syfu_H
    1 hour ago
















Why would you expect it to change anything in the vector? All you modify is an iterator

– UnholySheep
1 hour ago





Why would you expect it to change anything in the vector? All you modify is an iterator

– UnholySheep
1 hour ago




3




3





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
1 hour 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
1 hour ago














@RSahu: Here is the output from Ideone: ideone.com/19AVFF

– Syfu_H
1 hour ago





@RSahu: Here is the output from Ideone: ideone.com/19AVFF

– Syfu_H
1 hour ago












3 Answers
3






active

oldest

votes


















11














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)






share|improve this answer























  • What, no digression into lvalue qualified member functions?

    – Yakk - Adam Nevraumont
    1 hour ago


















5















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 cdagger typoan assign to begin()




Because



  1. Iterators are assignable.

  2. Rvalues of class type can be assigned .

  3. 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.






share|improve this answer

























  • 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
    1 hour 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
    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



















0














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.






share|improve this answer























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    3 Answers
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    3 Answers
    3






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    11














    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)






    share|improve this answer























    • What, no digression into lvalue qualified member functions?

      – Yakk - Adam Nevraumont
      1 hour ago















    11














    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)






    share|improve this answer























    • What, no digression into lvalue qualified member functions?

      – Yakk - Adam Nevraumont
      1 hour ago













    11












    11








    11







    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)






    share|improve this answer













    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)







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 1 hour ago









    BrianBrian

    66.2k798189




    66.2k798189












    • What, no digression into lvalue qualified member functions?

      – Yakk - Adam Nevraumont
      1 hour ago

















    • What, no digression into lvalue qualified member functions?

      – Yakk - Adam Nevraumont
      1 hour ago
















    What, no digression into lvalue qualified member functions?

    – Yakk - Adam Nevraumont
    1 hour ago





    What, no digression into lvalue qualified member functions?

    – Yakk - Adam Nevraumont
    1 hour ago













    5















    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 cdagger typoan assign to begin()




    Because



    1. Iterators are assignable.

    2. Rvalues of class type can be assigned .

    3. 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.






    share|improve this answer

























    • 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
      1 hour 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
      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
















    5















    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 cdagger typoan assign to begin()




    Because



    1. Iterators are assignable.

    2. Rvalues of class type can be assigned .

    3. 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.






    share|improve this answer

























    • 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
      1 hour 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
      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














    5












    5








    5








    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 cdagger typoan assign to begin()




    Because



    1. Iterators are assignable.

    2. Rvalues of class type can be assigned .

    3. 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.






    share|improve this answer
















    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 cdagger typoan assign to begin()




    Because



    1. Iterators are assignable.

    2. Rvalues of class type can be assigned .

    3. 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.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 13 mins ago

























    answered 1 hour 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
      1 hour 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
      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


















    • 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
      1 hour 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
      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

















    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
    1 hour 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
    1 hour 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
    1 hour 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
    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






    @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












    0














    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.






    share|improve this answer



























      0














      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.






      share|improve this answer

























        0












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        0







        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.






        share|improve this answer













        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.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 1 hour ago









        Nikos C.Nikos C.

        33.8k53967




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