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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 2 hours ago

bartop

asked 2 hours ago

bartopbartop

3,2331030

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  Lambdas do not participate in ADL
  
  – Guillaume Racicot
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  This isn't ADL. An int argument doesn't come from any namespace.
  
  – chris
  2 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
  1 hour 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 2 hours ago

bartop

asked 2 hours ago

bartopbartop

3,2331030

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  
  Lambdas do not participate in ADL
  
  – Guillaume Racicot
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  This isn't ADL. An int argument doesn't come from any namespace.
  
  – chris
  2 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
  1 hour 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 2 hours ago

bartop

asked 2 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 2 hours ago

bartop

asked 2 hours ago

bartopbartop

3,2331030

share|improve this question

edited 2 hours ago

bartop

asked 2 hours ago

bartopbartop

3,2331030

asked 2 hours ago

bartopbartop

3,2331030

asked 2 hours ago

bartopbartop

3,2331030

1 Answer
1

active

oldest

votes

3

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 5 mins ago

answered 20 mins ago

Michael KenzelMichael Kenzel

5,05811020

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  6 mins ago
  

  
  
  
  

add a comment | 

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3

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 5 mins ago

answered 20 mins ago

Michael KenzelMichael Kenzel

5,05811020

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  6 mins ago
  

  
  
  
  

add a comment | 

3

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 5 mins ago

answered 20 mins ago

Michael KenzelMichael Kenzel

5,05811020

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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
  6 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 5 mins ago

answered 20 mins ago

Michael KenzelMichael Kenzel

5,05811020

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 5 mins ago

answered 20 mins ago

Michael KenzelMichael Kenzel

5,05811020

answered 20 mins ago

Michael KenzelMichael Kenzel

5,05811020

answered 20 mins ago

Michael KenzelMichael Kenzel

5,05811020