Generic lambda vs generic function give different behaviour












13















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




















  • 5





    Lambdas do not participate in ADL

    – Guillaume Racicot
    12 hours ago






  • 8





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

    – chris
    12 hours ago






  • 4





    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
    12 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
    12 hours ago
















13















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




















  • 5





    Lambdas do not participate in ADL

    – Guillaume Racicot
    12 hours ago






  • 8





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

    – chris
    12 hours ago






  • 4





    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
    12 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
    12 hours ago














13












13








13


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






c++ lambda c++14







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




c++ lambda c++14






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 12 hours ago







bartop

















asked 12 hours ago









bartopbartop

3,2551031




3,2551031








  • 5





    Lambdas do not participate in ADL

    – Guillaume Racicot
    12 hours ago






  • 8





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

    – chris
    12 hours ago






  • 4





    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
    12 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
    12 hours ago














  • 5





    Lambdas do not participate in ADL

    – Guillaume Racicot
    12 hours ago






  • 8





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

    – chris
    12 hours ago






  • 4





    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
    12 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
    12 hours ago








5




5





Lambdas do not participate in ADL

– Guillaume Racicot
12 hours ago





Lambdas do not participate in ADL

– Guillaume Racicot
12 hours ago




8




8





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

– chris
12 hours ago





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

– chris
12 hours ago




4




4





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
12 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
12 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
12 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
12 hours ago












1 Answer
1






active

oldest

votes


















13














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
    10 hours ago











  • Seems like a right answet to me. If there is something that can be done to workaround my problem I would be thankful for comment/answer edit.

    – bartop
    4 hours ago






  • 1





    @Scheff indeed, I must've mixed up the URLs somehow. Should be fixed now. Thanks for pointing that out!

    – Michael Kenzel
    1 hour ago













  • @bartop can you not just remove the using directives and/or use fully-qualified names, e.g., baz::sort!?

    – Michael Kenzel
    5 mins ago













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1 Answer
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active

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1 Answer
1






active

oldest

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active

oldest

votes









13














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
    10 hours ago











  • Seems like a right answet to me. If there is something that can be done to workaround my problem I would be thankful for comment/answer edit.

    – bartop
    4 hours ago






  • 1





    @Scheff indeed, I must've mixed up the URLs somehow. Should be fixed now. Thanks for pointing that out!

    – Michael Kenzel
    1 hour ago













  • @bartop can you not just remove the using directives and/or use fully-qualified names, e.g., baz::sort!?

    – Michael Kenzel
    5 mins ago


















13














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
    10 hours ago











  • Seems like a right answet to me. If there is something that can be done to workaround my problem I would be thankful for comment/answer edit.

    – bartop
    4 hours ago






  • 1





    @Scheff indeed, I must've mixed up the URLs somehow. Should be fixed now. Thanks for pointing that out!

    – Michael Kenzel
    1 hour ago













  • @bartop can you not just remove the using directives and/or use fully-qualified names, e.g., baz::sort!?

    – Michael Kenzel
    5 mins ago
















13












13








13







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 1 hour ago

























answered 10 hours ago









Michael KenzelMichael Kenzel

5,19311121




5,19311121








  • 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
    10 hours ago











  • Seems like a right answet to me. If there is something that can be done to workaround my problem I would be thankful for comment/answer edit.

    – bartop
    4 hours ago






  • 1





    @Scheff indeed, I must've mixed up the URLs somehow. Should be fixed now. Thanks for pointing that out!

    – Michael Kenzel
    1 hour ago













  • @bartop can you not just remove the using directives and/or use fully-qualified names, e.g., baz::sort!?

    – Michael Kenzel
    5 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
    10 hours ago











  • Seems like a right answet to me. If there is something that can be done to workaround my problem I would be thankful for comment/answer edit.

    – bartop
    4 hours ago






  • 1





    @Scheff indeed, I must've mixed up the URLs somehow. Should be fixed now. Thanks for pointing that out!

    – Michael Kenzel
    1 hour ago













  • @bartop can you not just remove the using directives and/or use fully-qualified names, e.g., baz::sort!?

    – Michael Kenzel
    5 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
10 hours 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
10 hours ago













Seems like a right answet to me. If there is something that can be done to workaround my problem I would be thankful for comment/answer edit.

– bartop
4 hours ago





Seems like a right answet to me. If there is something that can be done to workaround my problem I would be thankful for comment/answer edit.

– bartop
4 hours ago




1




1





@Scheff indeed, I must've mixed up the URLs somehow. Should be fixed now. Thanks for pointing that out!

– Michael Kenzel
1 hour ago







@Scheff indeed, I must've mixed up the URLs somehow. Should be fixed now. Thanks for pointing that out!

– Michael Kenzel
1 hour ago















@bartop can you not just remove the using directives and/or use fully-qualified names, e.g., baz::sort!?

– Michael Kenzel
5 mins ago







@bartop can you not just remove the using directives and/or use fully-qualified names, e.g., baz::sort!?

– Michael Kenzel
5 mins ago






















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