Can a lambda expression be passed as function poin

Question

I am trying to pass a lambda expression to a function that takes a function pointer, is this even possible?

Here is some sample code, I'm using VS2010:

#include <iostream>
using namespace std;

void func(int i){cout << "I'V BEEN CALLED: " << i <<endl;}

void fptrfunc(void (*fptr)(int i), int j){fptr(j);}

int main(){
    fptrfunc(func,10); //this is ok
    fptrfunc([](int i){cout << "LAMBDA CALL " << i << endl; }, 20); //DOES NOT COMPILE
    return 0;
}

Answer 1

This works in VS2010:

template<class FunctorT>
void* getcodeptr(const FunctorT& f) {
  auto ptr = &FunctorT::operator();
  return *(void**)&ptr;
} 

void main() {
  auto hello = [](char* name){ printf("hello %s\n", name); };
  void(*pfn)(char*) = (void(*)(char*)) getcodeptr(hello);
  pfn("world");
}

Answer 2

In VC10 RTM, no - but after the lambda feature in VC10 was finalized, the standard committee did add language which allows stateless lambdas to degrade to function pointers. So in the future this will be possible.

Answer 3

You can use std::function for this:

void fptrfunc(std::function<void (int)> fun, int j)
{
    fun(j);
}

Or go completely generic:

template <typename Fun>
void fptrfunc(Fun fun, int j)
{
    fun(j);
}

Answer 4

No. It cannot. Not dependably at least. I know that VS2010 implements them as object functors. Based on how they work that may be an a priori requirement.

Answer 5

As long as the lambda doesn't use capture clause (i.e. doesn't capture variables from above it), it can be used as a function pointer. VC compiler internally generates anonymous functions with different calling convention so that it can be used without issues.