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Generic lambda vs generic function give different behaviour


What is a lambda (function)?What is the difference between #include <filename> and #include “filename”?What are the differences between a pointer variable and a reference variable in C++?What is the difference between a 'closure' and a 'lambda'?Why are Python lambdas useful?Distinct() with lambda?list comprehension vs. lambda + filterWhat is a lambda expression in C++11?Calling `this` member function from generic lambda - clang vs gccConstructing std::function argument from lambda













10















Take following code as an example



#include <algorithm>

namespace baz
template<class T>
void sort(T&&)


namespace boot
const auto sort = [](auto &&);


void foo ()
using namespace std;
using namespace baz;
sort(1);


void bar()
using namespace std;
using namespace boot;
sort(1);



I expected that since foo compiled, then bar shall compile as well. To my surprise, the foo compiles correctly and bar has problem with ambiguous call to sort function. Am I doing something illegal here or this is proper way compiler should behave? If so, why is it so different. I though generic lambda can be treated as syntactic sugar for generic function.



live example










share|improve this question



















  • 4





    Lambdas do not participate in ADL

    – Guillaume Racicot
    3 hours ago






  • 5





    This isn't ADL. An int argument doesn't come from any namespace.

    – chris
    3 hours ago






  • 2





    Should this really be ambiguous, though? std::sort() doesn't take 1 parameter as input, it takes at least 2, so why is the compiler even considering it as a candidate for a call that passes only 1 parameter value?

    – Remy Lebeau
    2 hours ago












  • There must be something about the extra layer of indirection that the lambda introduces. With the first example, the call is made to ::baz::sort, but in the second example, it would have to find ::boot::mystery_lambda_type::operator(). That extra step might be what causes std::sort to be considered first. I don't have the standard in front of me so can't be sure about this.

    – alter igel
    2 hours ago















10















Take following code as an example



#include <algorithm>

namespace baz
template<class T>
void sort(T&&)


namespace boot
const auto sort = [](auto &&);


void foo ()
using namespace std;
using namespace baz;
sort(1);


void bar()
using namespace std;
using namespace boot;
sort(1);



I expected that since foo compiled, then bar shall compile as well. To my surprise, the foo compiles correctly and bar has problem with ambiguous call to sort function. Am I doing something illegal here or this is proper way compiler should behave? If so, why is it so different. I though generic lambda can be treated as syntactic sugar for generic function.



live example










share|improve this question



















  • 4





    Lambdas do not participate in ADL

    – Guillaume Racicot
    3 hours ago






  • 5





    This isn't ADL. An int argument doesn't come from any namespace.

    – chris
    3 hours ago






  • 2





    Should this really be ambiguous, though? std::sort() doesn't take 1 parameter as input, it takes at least 2, so why is the compiler even considering it as a candidate for a call that passes only 1 parameter value?

    – Remy Lebeau
    2 hours ago












  • There must be something about the extra layer of indirection that the lambda introduces. With the first example, the call is made to ::baz::sort, but in the second example, it would have to find ::boot::mystery_lambda_type::operator(). That extra step might be what causes std::sort to be considered first. I don't have the standard in front of me so can't be sure about this.

    – alter igel
    2 hours ago













10












10








10


1






Take following code as an example



#include <algorithm>

namespace baz
template<class T>
void sort(T&&)


namespace boot
const auto sort = [](auto &&);


void foo ()
using namespace std;
using namespace baz;
sort(1);


void bar()
using namespace std;
using namespace boot;
sort(1);



I expected that since foo compiled, then bar shall compile as well. To my surprise, the foo compiles correctly and bar has problem with ambiguous call to sort function. Am I doing something illegal here or this is proper way compiler should behave? If so, why is it so different. I though generic lambda can be treated as syntactic sugar for generic function.



live example










share|improve this question
















Take following code as an example



#include <algorithm>

namespace baz
template<class T>
void sort(T&&)


namespace boot
const auto sort = [](auto &&);


void foo ()
using namespace std;
using namespace baz;
sort(1);


void bar()
using namespace std;
using namespace boot;
sort(1);



I expected that since foo compiled, then bar shall compile as well. To my surprise, the foo compiles correctly and bar has problem with ambiguous call to sort function. Am I doing something illegal here or this is proper way compiler should behave? If so, why is it so different. I though generic lambda can be treated as syntactic sugar for generic function.



live example







c++ lambda c++14






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 3 hours ago







bartop

















asked 3 hours ago









bartopbartop

3,2381030




3,2381030







  • 4





    Lambdas do not participate in ADL

    – Guillaume Racicot
    3 hours ago






  • 5





    This isn't ADL. An int argument doesn't come from any namespace.

    – chris
    3 hours ago






  • 2





    Should this really be ambiguous, though? std::sort() doesn't take 1 parameter as input, it takes at least 2, so why is the compiler even considering it as a candidate for a call that passes only 1 parameter value?

    – Remy Lebeau
    2 hours ago












  • There must be something about the extra layer of indirection that the lambda introduces. With the first example, the call is made to ::baz::sort, but in the second example, it would have to find ::boot::mystery_lambda_type::operator(). That extra step might be what causes std::sort to be considered first. I don't have the standard in front of me so can't be sure about this.

    – alter igel
    2 hours ago












  • 4





    Lambdas do not participate in ADL

    – Guillaume Racicot
    3 hours ago






  • 5





    This isn't ADL. An int argument doesn't come from any namespace.

    – chris
    3 hours ago






  • 2





    Should this really be ambiguous, though? std::sort() doesn't take 1 parameter as input, it takes at least 2, so why is the compiler even considering it as a candidate for a call that passes only 1 parameter value?

    – Remy Lebeau
    2 hours ago












  • There must be something about the extra layer of indirection that the lambda introduces. With the first example, the call is made to ::baz::sort, but in the second example, it would have to find ::boot::mystery_lambda_type::operator(). That extra step might be what causes std::sort to be considered first. I don't have the standard in front of me so can't be sure about this.

    – alter igel
    2 hours ago







4




4





Lambdas do not participate in ADL

– Guillaume Racicot
3 hours ago





Lambdas do not participate in ADL

– Guillaume Racicot
3 hours ago




5




5





This isn't ADL. An int argument doesn't come from any namespace.

– chris
3 hours ago





This isn't ADL. An int argument doesn't come from any namespace.

– chris
3 hours ago




2




2





Should this really be ambiguous, though? std::sort() doesn't take 1 parameter as input, it takes at least 2, so why is the compiler even considering it as a candidate for a call that passes only 1 parameter value?

– Remy Lebeau
2 hours ago






Should this really be ambiguous, though? std::sort() doesn't take 1 parameter as input, it takes at least 2, so why is the compiler even considering it as a candidate for a call that passes only 1 parameter value?

– Remy Lebeau
2 hours ago














There must be something about the extra layer of indirection that the lambda introduces. With the first example, the call is made to ::baz::sort, but in the second example, it would have to find ::boot::mystery_lambda_type::operator(). That extra step might be what causes std::sort to be considered first. I don't have the standard in front of me so can't be sure about this.

– alter igel
2 hours ago





There must be something about the extra layer of indirection that the lambda introduces. With the first example, the call is made to ::baz::sort, but in the second example, it would have to find ::boot::mystery_lambda_type::operator(). That extra step might be what causes std::sort to be considered first. I don't have the standard in front of me so can't be sure about this.

– alter igel
2 hours ago












2 Answers
2






active

oldest

votes


















7














The problem here is not that the call to sort is ambiguous, but that the name sort is ambiguous. Name lookup happens before overload resolution.



I believe the relevant section is [basic.lookup]/1, specifically




[…] The declarations found by name lookup shall either all denote the same entity or shall all denote functions or function templates; in the latter case, the declarations are said to form a set of overloaded functions ([over.load]). […]




In your case, the name sort denotes both, the object boot::sort as well as the set of overloaded functions std::sort. Therefore, name lookup fails.



Your code is really no different from if you had written, for example



namespace baz 
int a;


namespace boot
int a;


void foo()
using namespace baz;
using namespace boot;
a = 42; // error: reference to 'a' is ambiguous



Try it out here; compare this to a case that actually has an ambiguous function call; note how the error message is the same as in your case, specifically referring to the name itself being ambiguous rather than the function call.






share|improve this answer




















  • 1





    I think this is actually the right answer. And I would like to add that if both the template function sort and the lambda sort were declared in the same namespace, it would be an error. You cannot have a function and non-function with the same name in the same namespace. So there could never be an overload set that has both true functions and function-like objects.

    – Mike
    58 mins ago











  • @Mike Kudos to you, Sir, for deleting your answer despite putting quite some effort into it and getting that many upvotes!

    – Michael Kenzel
    35 mins ago


















0














I used the function typeid to watch how compiler identify lambda, function, function pointer, std::function, and each in them is different from any other.



I have not clarified why compiler work in this way, but obviously, behavior compiled between lambda and function are different.lambda must check it has the authority to use or change which value declared in this namespace when compiling, so the range compiled is limited, situation that can use all just like a function is precious few. At least there is essential distinction between them.



Consider about when lambda is used anonymously, compiler may not have to remember the initial address of lambda, compare lambda is impossible.I thought that's one of reasons.



And there should be more practical reasons for compiler has no support for distinguish named lambda from function by parameter list or return type, if it appeared in future, it won't be named overload probably.






share|improve this answer










New contributor




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



















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7














    The problem here is not that the call to sort is ambiguous, but that the name sort is ambiguous. Name lookup happens before overload resolution.



    I believe the relevant section is [basic.lookup]/1, specifically




    […] The declarations found by name lookup shall either all denote the same entity or shall all denote functions or function templates; in the latter case, the declarations are said to form a set of overloaded functions ([over.load]). […]




    In your case, the name sort denotes both, the object boot::sort as well as the set of overloaded functions std::sort. Therefore, name lookup fails.



    Your code is really no different from if you had written, for example



    namespace baz 
    int a;


    namespace boot
    int a;


    void foo()
    using namespace baz;
    using namespace boot;
    a = 42; // error: reference to 'a' is ambiguous



    Try it out here; compare this to a case that actually has an ambiguous function call; note how the error message is the same as in your case, specifically referring to the name itself being ambiguous rather than the function call.






    share|improve this answer




















    • 1





      I think this is actually the right answer. And I would like to add that if both the template function sort and the lambda sort were declared in the same namespace, it would be an error. You cannot have a function and non-function with the same name in the same namespace. So there could never be an overload set that has both true functions and function-like objects.

      – Mike
      58 mins ago











    • @Mike Kudos to you, Sir, for deleting your answer despite putting quite some effort into it and getting that many upvotes!

      – Michael Kenzel
      35 mins ago















    7














    The problem here is not that the call to sort is ambiguous, but that the name sort is ambiguous. Name lookup happens before overload resolution.



    I believe the relevant section is [basic.lookup]/1, specifically




    […] The declarations found by name lookup shall either all denote the same entity or shall all denote functions or function templates; in the latter case, the declarations are said to form a set of overloaded functions ([over.load]). […]




    In your case, the name sort denotes both, the object boot::sort as well as the set of overloaded functions std::sort. Therefore, name lookup fails.



    Your code is really no different from if you had written, for example



    namespace baz 
    int a;


    namespace boot
    int a;


    void foo()
    using namespace baz;
    using namespace boot;
    a = 42; // error: reference to 'a' is ambiguous



    Try it out here; compare this to a case that actually has an ambiguous function call; note how the error message is the same as in your case, specifically referring to the name itself being ambiguous rather than the function call.






    share|improve this answer




















    • 1





      I think this is actually the right answer. And I would like to add that if both the template function sort and the lambda sort were declared in the same namespace, it would be an error. You cannot have a function and non-function with the same name in the same namespace. So there could never be an overload set that has both true functions and function-like objects.

      – Mike
      58 mins ago











    • @Mike Kudos to you, Sir, for deleting your answer despite putting quite some effort into it and getting that many upvotes!

      – Michael Kenzel
      35 mins ago













    7












    7








    7







    The problem here is not that the call to sort is ambiguous, but that the name sort is ambiguous. Name lookup happens before overload resolution.



    I believe the relevant section is [basic.lookup]/1, specifically




    […] The declarations found by name lookup shall either all denote the same entity or shall all denote functions or function templates; in the latter case, the declarations are said to form a set of overloaded functions ([over.load]). […]




    In your case, the name sort denotes both, the object boot::sort as well as the set of overloaded functions std::sort. Therefore, name lookup fails.



    Your code is really no different from if you had written, for example



    namespace baz 
    int a;


    namespace boot
    int a;


    void foo()
    using namespace baz;
    using namespace boot;
    a = 42; // error: reference to 'a' is ambiguous



    Try it out here; compare this to a case that actually has an ambiguous function call; note how the error message is the same as in your case, specifically referring to the name itself being ambiguous rather than the function call.






    share|improve this answer















    The problem here is not that the call to sort is ambiguous, but that the name sort is ambiguous. Name lookup happens before overload resolution.



    I believe the relevant section is [basic.lookup]/1, specifically




    […] The declarations found by name lookup shall either all denote the same entity or shall all denote functions or function templates; in the latter case, the declarations are said to form a set of overloaded functions ([over.load]). […]




    In your case, the name sort denotes both, the object boot::sort as well as the set of overloaded functions std::sort. Therefore, name lookup fails.



    Your code is really no different from if you had written, for example



    namespace baz 
    int a;


    namespace boot
    int a;


    void foo()
    using namespace baz;
    using namespace boot;
    a = 42; // error: reference to 'a' is ambiguous



    Try it out here; compare this to a case that actually has an ambiguous function call; note how the error message is the same as in your case, specifically referring to the name itself being ambiguous rather than the function call.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 56 mins ago

























    answered 1 hour ago









    Michael KenzelMichael Kenzel

    5,09811020




    5,09811020







    • 1





      I think this is actually the right answer. And I would like to add that if both the template function sort and the lambda sort were declared in the same namespace, it would be an error. You cannot have a function and non-function with the same name in the same namespace. So there could never be an overload set that has both true functions and function-like objects.

      – Mike
      58 mins ago











    • @Mike Kudos to you, Sir, for deleting your answer despite putting quite some effort into it and getting that many upvotes!

      – Michael Kenzel
      35 mins ago












    • 1





      I think this is actually the right answer. And I would like to add that if both the template function sort and the lambda sort were declared in the same namespace, it would be an error. You cannot have a function and non-function with the same name in the same namespace. So there could never be an overload set that has both true functions and function-like objects.

      – Mike
      58 mins ago











    • @Mike Kudos to you, Sir, for deleting your answer despite putting quite some effort into it and getting that many upvotes!

      – Michael Kenzel
      35 mins ago







    1




    1





    I think this is actually the right answer. And I would like to add that if both the template function sort and the lambda sort were declared in the same namespace, it would be an error. You cannot have a function and non-function with the same name in the same namespace. So there could never be an overload set that has both true functions and function-like objects.

    – Mike
    58 mins ago





    I think this is actually the right answer. And I would like to add that if both the template function sort and the lambda sort were declared in the same namespace, it would be an error. You cannot have a function and non-function with the same name in the same namespace. So there could never be an overload set that has both true functions and function-like objects.

    – Mike
    58 mins ago













    @Mike Kudos to you, Sir, for deleting your answer despite putting quite some effort into it and getting that many upvotes!

    – Michael Kenzel
    35 mins ago





    @Mike Kudos to you, Sir, for deleting your answer despite putting quite some effort into it and getting that many upvotes!

    – Michael Kenzel
    35 mins ago













    0














    I used the function typeid to watch how compiler identify lambda, function, function pointer, std::function, and each in them is different from any other.



    I have not clarified why compiler work in this way, but obviously, behavior compiled between lambda and function are different.lambda must check it has the authority to use or change which value declared in this namespace when compiling, so the range compiled is limited, situation that can use all just like a function is precious few. At least there is essential distinction between them.



    Consider about when lambda is used anonymously, compiler may not have to remember the initial address of lambda, compare lambda is impossible.I thought that's one of reasons.



    And there should be more practical reasons for compiler has no support for distinguish named lambda from function by parameter list or return type, if it appeared in future, it won't be named overload probably.






    share|improve this answer










    New contributor




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
























      0














      I used the function typeid to watch how compiler identify lambda, function, function pointer, std::function, and each in them is different from any other.



      I have not clarified why compiler work in this way, but obviously, behavior compiled between lambda and function are different.lambda must check it has the authority to use or change which value declared in this namespace when compiling, so the range compiled is limited, situation that can use all just like a function is precious few. At least there is essential distinction between them.



      Consider about when lambda is used anonymously, compiler may not have to remember the initial address of lambda, compare lambda is impossible.I thought that's one of reasons.



      And there should be more practical reasons for compiler has no support for distinguish named lambda from function by parameter list or return type, if it appeared in future, it won't be named overload probably.






      share|improve this answer










      New contributor




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






















        0












        0








        0







        I used the function typeid to watch how compiler identify lambda, function, function pointer, std::function, and each in them is different from any other.



        I have not clarified why compiler work in this way, but obviously, behavior compiled between lambda and function are different.lambda must check it has the authority to use or change which value declared in this namespace when compiling, so the range compiled is limited, situation that can use all just like a function is precious few. At least there is essential distinction between them.



        Consider about when lambda is used anonymously, compiler may not have to remember the initial address of lambda, compare lambda is impossible.I thought that's one of reasons.



        And there should be more practical reasons for compiler has no support for distinguish named lambda from function by parameter list or return type, if it appeared in future, it won't be named overload probably.






        share|improve this answer










        New contributor




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










        I used the function typeid to watch how compiler identify lambda, function, function pointer, std::function, and each in them is different from any other.



        I have not clarified why compiler work in this way, but obviously, behavior compiled between lambda and function are different.lambda must check it has the authority to use or change which value declared in this namespace when compiling, so the range compiled is limited, situation that can use all just like a function is precious few. At least there is essential distinction between them.



        Consider about when lambda is used anonymously, compiler may not have to remember the initial address of lambda, compare lambda is impossible.I thought that's one of reasons.



        And there should be more practical reasons for compiler has no support for distinguish named lambda from function by parameter list or return type, if it appeared in future, it won't be named overload probably.







        share|improve this answer










        New contributor




        LuLi 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 answer



        share|improve this answer








        edited 22 mins ago





















        New contributor




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









        answered 30 mins ago









        LuLiLuLi

        233




        233




        New contributor




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





        New contributor





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






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



























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