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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

9

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

edited 3 hours ago

bartop

asked 3 hours ago

bartopbartop

3,2331030

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

9

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

edited 3 hours ago

bartop

asked 3 hours ago

bartopbartop

3,2331030

• 
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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

edited 3 hours ago

bartop

asked 3 hours ago

bartopbartop

3,2331030

#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);

}

share|improve this question

edited 3 hours ago

bartop

asked 3 hours ago

bartopbartop

3,2331030

share|improve this question

edited 3 hours ago

bartop

asked 3 hours ago

bartopbartop

3,2331030

asked 3 hours ago

bartopbartop

3,2331030

asked 3 hours ago

bartopbartop

3,2331030

2 Answers
2

active

oldest

votes

4

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

edited 47 mins ago

answered 1 hour ago

Michael KenzelMichael Kenzel

5,06811020

• 
  
  
  

  
  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
  48 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
  25 mins ago
  

  
  
  
  

add a comment | 

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

edited 12 mins ago

answered 20 mins ago

LuLiLuLi

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.

add a comment | 

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4

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

edited 47 mins ago

answered 1 hour ago

Michael KenzelMichael Kenzel

5,06811020

• 
  
  
  

  
  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
  48 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
  25 mins ago
  

  
  
  
  

add a comment | 

4

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

edited 47 mins ago

answered 1 hour ago

Michael KenzelMichael Kenzel

5,06811020

• 
  
  
  

  
  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
  48 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
  25 mins ago
  

  
  
  
  

add a comment | 

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

edited 47 mins ago

answered 1 hour ago

Michael KenzelMichael Kenzel

5,06811020

namespace baz {

    int a;

}



namespace boot {

    int a;

}



void foo() {

    using namespace baz;

    using namespace boot;

    a = 42;  // error: reference to 'a' is ambiguous

}

share|improve this answer

edited 47 mins ago

answered 1 hour ago

Michael KenzelMichael Kenzel

5,06811020

answered 1 hour ago

Michael KenzelMichael Kenzel

5,06811020

answered 1 hour ago

Michael KenzelMichael Kenzel

5,06811020

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

edited 12 mins ago

answered 20 mins ago

LuLiLuLi

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.

add a comment | 

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

edited 12 mins ago

answered 20 mins ago

LuLiLuLi

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.

add a comment | 

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

edited 12 mins ago

answered 20 mins ago

LuLiLuLi

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.

share|improve this answer

edited 12 mins ago

answered 20 mins ago

LuLiLuLi

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.

answered 20 mins ago

LuLiLuLi

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.

answered 20 mins ago

LuLiLuLi

233