CODEWITHSUNDEEP
c++c++17
template boy
template boy

Reputation: 10480

Lambda as default argument fails

I get an error with the latest versions of clang and gcc with this code:

int main() {
    auto lambda = [] (auto = [] {}) {};
    lambda();
}

Clang gives the error:

prog.cc: In function 'int main()':
prog.cc:3:12: error: no match for call to '(main()::<lambda(auto:1)>) ()'
     lambda();
            ^
prog.cc:2:35: note: candidate: template<class auto:1> main()::<lambda(auto:1)>
     auto lambda = [] (auto = [] {}) {};
                                   ^
prog.cc:2:35: note:   template argument deduction/substitution failed:
prog.cc:3:12: note:   couldn't deduce template parameter 'auto:1'
     lambda();
            ^

Why does this fail?

Upvotes: 19

Views: 1363

Answers (3)

Denilson Amorim
Denilson Amorim

Reputation: 10292

Since lambdas are sugar for functors, the issue is on the fact template functions are unable to deduce template arguments (auto) in this default context.

A lambda can be reduced to the functor struct level by taking in consideration those statements:

§5.1.2/3 [expr.prim.lambda]

The type of the lambda-expression (which is also the type of the closure object) is a unique, unnamed non-union class type [...]

§5.1.2/5 [expr.prim.lambda]

[...] For a generic lambda, the closure type has a public inline function call operator member template (14.5.2) whose template-parameter-list consists of one invented type template-parameter for each occurrence of auto in the lambda’s parameter-declaration-clause, in order of appearance. [...]

As such the type of your lambda is equivalent to this functor type:

struct unnamed
{
    template<typename Auto1>
    auto operator()(Auto1 = []{})
    {
    }
};

And your usage is then equivalent to:

int main() {
    auto lambda = unnamed();
    lambda();
}

The type of Auto1 is unable to be inferred in this context as specified in §14.8.2.5/5 [temp.deduct.type]:

The non-deduced contexts are:

[...]

— A template parameter used in the parameter type of a function parameter that has a default argument that is being used in the call for which argument deduction is being done.

Upvotes: 13

Yakk - Adam Nevraumont
Yakk - Adam Nevraumont

Reputation: 275720

Template functions (or methods) do not deduce their type parameters from their default arguments, and a closure with auto parameters is merely an object with a template method.

This makes having a default lambda for a template function a bit annoying.

One approach would be to type erase calling an object, without storing it, like so:

#include <utility>
#include <type_traits>
#include <memory>

template<class Sig>
struct function_view;

template<class R, class...Args>
struct function_view<R(Args...)>{
  void* state;
  R(*f)(void*, Args&&...);

  template<class F, class=std::enable_if_t<std::is_convertible<std::result_of_t<F&(Args...)>,R>{}>>
  function_view( F&& fin ):
    state(const_cast<void*>(static_cast<void*>(std::addressof(fin)))),
    f( [](void* state, Args&&...args)->R{
      F&& f = std::forward<F>(*static_cast<std::decay_t<F>*>(state));
      return f(std::forward<Args>(args)...);
    })
  {}
  function_view( R(*fin)(Args...) ):
    state(fin),
    f( fin?+[](void* state, Args&&...args)->R{
      R(*f)(Args...) = static_cast<R(*)(Args...)>(state);
      return f(std::forward<Args>(args)...);
    }:nullptr)
  {}
  explicit operator bool(){return f;}
  function_view():state(nullptr),f(nullptr){}
  function_view(std::nullptr_t):function_view(){}
  R operator()(Args...args)const{
    return f(state, std::forward<Args>(args)...);
  }
};
template<class...Args>
struct function_view<void(Args...)>{
  void* state;
  void(*f)(void*, Args&&...);

  template<class F, class=std::result_of_t<F&(Args...)>>
  function_view( F&& fin ):
    state(const_cast<void*>(static_cast<void*>(std::addressof(fin)))),
    f( [](void* state, Args&&...args){
      F&& f = std::forward<F>(*static_cast<std::decay_t<F>*>(state));
      f(std::forward<Args>(args)...);
    })
  {}
  function_view( void(*fin)(Args...) ):
    state(fin),
    f( fin?+[](void* state, Args&&...args){
      void(*f)(Args...) = static_cast<void(*)(Args...)>(state);
      f(std::forward<Args>(args)...);
    }:nullptr)
  {}

  explicit operator bool(){return f;}
  function_view():state(nullptr),f(nullptr){}
  function_view(std::nullptr_t):function_view(){}
  void operator()(Args...args)const{
    f(state, std::forward<Args>(args)...);
  }
};

int main() {
  auto f = [] (function_view<void()> x=[]{}) {
    x();
  };
  f();
}

As this just works with function pointers, and I have had good experience with gcc inlining simple function pointers, it might not have as high a performance impact as std::function. And unlike std::function no virtual tables or heap allocation is involved.

live example

For a non-lambda, you can do this:

template<class X=function_view<void()>>
void f( X&& x=[]{} ) {
  x();
}

which deduces if you pass is an argument, and becomes a function-at-nothing if you don't. You could also do:

struct do_nothing {
  template<class...Args>
  void operator()(Args&&...)const{}
};

template<class X=do_nothing>
void f( X&& x=do_nothing{} ) {
  x();
}

which might be easier to optimize.

Upvotes: 2

Daniel Frey
Daniel Frey

Reputation: 56883

Type deduction for auto does not consider default arguments.

Upvotes: 22

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