[FIXED] What is the type of lambda when deduced with "auto" in C++11?

Issue

I had a perception that, type of a lambda is a function pointer. When I performed following test, I found it to be wrong (demo).

#define LAMBDA [] (int i) -> long { return 0; }
int main ()
{
  long (*pFptr)(int) = LAMBDA;  // ok
  auto pAuto = LAMBDA;  // ok
  assert(typeid(pFptr) == typeid(pAuto));  // assertion fails !
}

Is above code missing any point ? If not then, what is the typeof a lambda expression when deduced with auto keyword ?

Solution

The type of a lambda expression is unspecified.

But they are generally mere syntactic sugar for functors. A lambda is translated directly into a functor. Anything inside the [] are turned into constructor parameters and members of the functor object, and the parameters inside () are turned into parameters for the functor's operator().

A lambda which captures no variables (nothing inside the []'s) can be converted into a function pointer (MSVC2010 doesn't support this, if that's your compiler, but this conversion is part of the standard).

But the actual type of the lambda isn't a function pointer. It's some unspecified functor type.

Answered By - jalf

Answer Checked By - David Marino (PHPFixing Volunteer)