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”?

Man transported from Alternate World into ours by a Neutrino Detector

"Eavesdropping" vs "Listen in on"

What would be the main consequences for a country leaving the WTO?

TikZ: How to fill area with a special pattern?

What does "shotgun unity" refer to here in this sentence?

Is it correct to say moon starry nights?

IC has pull-down resistors on SMBus lines?

Why do we say 'Un seul M' and not 'Une seule M' even though M is a "consonne"

Is it OK to decorate a log book cover?

Calculate the Mean mean of two numbers

Where do students learn to solve polynomial equations these days?

what's the use of '% to gdp' type of variables?

Why am I getting "Static method cannot be referenced from a non static context: String String.valueOf(Object)"?

Is there such a thing as a proper verb, like a proper noun?

Is there a way to save my career from absolute disaster?

How do I fit a non linear curve?

Reshaping json / reparing json inside shell script (remove trailing comma)

Lucky Feat: How can "more than one creature spend a luck point to influence the outcome of a roll"?

free fall ellipse or parabola?

How to find image of a complex function with given constraints?

Is a distribution that is normal, but highly skewed, considered Gaussian?

The Ultimate Number Sequence Puzzle

Vector calculus integration identity problem

What day is it again?



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”?










8















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.















  • 2





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

    – UnholySheep
    2 days ago






  • 7





    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 days ago












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

    – Syfu_H
    2 days ago






  • 1





    Why would anyone want to close a clear question that touches a non trivial C++ subject?

    – curiousguy
    2 days ago















8















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.















  • 2





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

    – UnholySheep
    2 days ago






  • 7





    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 days ago












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

    – Syfu_H
    2 days ago






  • 1





    Why would anyone want to close a clear question that touches a non trivial C++ subject?

    – curiousguy
    2 days ago













8












8








8


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 2 days ago









Neil Butterworth

27.3k54681




27.3k54681






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 2 days ago









Syfu_HSyfu_H

744




744




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.







  • 2





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

    – UnholySheep
    2 days ago






  • 7





    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 days ago












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

    – Syfu_H
    2 days ago






  • 1





    Why would anyone want to close a clear question that touches a non trivial C++ subject?

    – curiousguy
    2 days ago












  • 2





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

    – UnholySheep
    2 days ago






  • 7





    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 days ago












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

    – Syfu_H
    2 days ago






  • 1





    Why would anyone want to close a clear question that touches a non trivial C++ subject?

    – curiousguy
    2 days ago







2




2





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

– UnholySheep
2 days ago





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

– UnholySheep
2 days ago




7




7





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 days 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 days ago














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

– Syfu_H
2 days ago





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

– Syfu_H
2 days ago




1




1





Why would anyone want to close a clear question that touches a non trivial C++ subject?

– curiousguy
2 days ago





Why would anyone want to close a clear question that touches a non trivial C++ subject?

– curiousguy
2 days ago












3 Answers
3






active

oldest

votes


















17














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


















  • 4





    What, no digression into lvalue qualified member functions?

    – Yakk - Adam Nevraumont
    2 days ago


















6















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



  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
    2 days 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 days 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
    2 days ago



















3














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























    Your Answer






    StackExchange.ifUsing("editor", function ()
    StackExchange.using("externalEditor", function ()
    StackExchange.using("snippets", function ()
    StackExchange.snippets.init();
    );
    );
    , "code-snippets");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "1"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );






    Syfu_H is a new contributor. Be nice, and check out our Code of Conduct.









    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55424362%2fwhat-is-the-result-of-assigning-to-stdvectortbegin%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    17














    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


















    • 4





      What, no digression into lvalue qualified member functions?

      – Yakk - Adam Nevraumont
      2 days ago















    17














    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


















    • 4





      What, no digression into lvalue qualified member functions?

      – Yakk - Adam Nevraumont
      2 days ago













    17












    17








    17







    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 2 days ago









    BrianBrian

    66.3k798190




    66.3k798190







    • 4





      What, no digression into lvalue qualified member functions?

      – Yakk - Adam Nevraumont
      2 days ago












    • 4





      What, no digression into lvalue qualified member functions?

      – Yakk - Adam Nevraumont
      2 days ago







    4




    4





    What, no digression into lvalue qualified member functions?

    – Yakk - Adam Nevraumont
    2 days ago





    What, no digression into lvalue qualified member functions?

    – Yakk - Adam Nevraumont
    2 days ago













    6















    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



    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
      2 days 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 days 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
      2 days ago
















    6















    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



    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
      2 days 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 days 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
      2 days ago














    6












    6








    6








    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



    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 can 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 2 days ago

























    answered 2 days ago









    eerorikaeerorika

    88.4k663135




    88.4k663135












    • 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 days 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 days 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
      2 days 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 days 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 days 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
      2 days 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 days 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 days 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 days 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 days 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
    2 days 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
    2 days ago












    3














    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



























      3














      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

























        3












        3








        3







        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 2 days ago









        Nikos C.Nikos C.

        33.9k53967




        33.9k53967




















            Syfu_H is a new contributor. Be nice, and check out our Code of Conduct.









            draft saved

            draft discarded


















            Syfu_H is a new contributor. Be nice, and check out our Code of Conduct.












            Syfu_H is a new contributor. Be nice, and check out our Code of Conduct.











            Syfu_H is a new contributor. Be nice, and check out our Code of Conduct.














            Thanks for contributing an answer to Stack Overflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55424362%2fwhat-is-the-result-of-assigning-to-stdvectortbegin%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            រឿង រ៉ូមេអូ និង ហ្ស៊ុយលីយេ សង្ខេបរឿង តួអង្គ បញ្ជីណែនាំ

            QGIS export composer to PDF scale the map [closed] Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Print Composer QGIS 2.6, how to export image?QGIS 2.8.1 print composer won't export all OpenCycleMap base layer tilesSave Print/Map QGIS composer view as PNG/PDF using Python (without changing anything in visible layout)?Export QGIS Print Composer PDF with searchable text labelsQGIS Print Composer does not change from landscape to portrait orientation?How can I avoid map size and scale changes in print composer?Fuzzy PDF export in QGIS running on macSierra OSExport the legend into its 100% size using Print ComposerScale-dependent rendering in QGIS PDF output

            PDF-ში გადმოწერა სანავიგაციო მენიუproject page