c++ - automatic decay of lambda to function pointer when passing to template function

Question

Is there a way to make a lambda decay to a pointer, without explicitly casting to the right signature? This would tidy some code:

template<typename T> T call(T(*func)()){ return func(); }
int ptr(){ return 0; }
int main(){
    auto ret1 = call(ptr);
    auto ret2 = call((int(*)())([]{ return 0; }));
    auto ret3 = call([]{ return 0; });  //won't compile
}

It's evident that a call to call works only if the lambda decays to a pointer, but I'm guessing that that can happen only after the right function overload/template is chosen. Unfortunately I can only think of solutions that involve templates to make a lambda with any signature decay, so I'm back to square one.

Answer 1

You can change your lambda to use the unary + operator: +[]{ return 0; }

This works because unary plus can be applied to pointers, and will trigger the implicit conversion to function pointer.