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 ?
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.