Function default parameters are ignored

Question

For this simplified piece of code I'm getting following error:

error: too few arguments to function std::cout << f();

int g(int a = 2, int b = 1)
{
    return a + b;
}

template<class Func>
void generic(Func f)
{
    std::cout << f();
}

int main()
{
    generic(g);
}

I cannot clue the reason for why the default parameters of function f are not passing into a function generic. It behaves like f doesn't have any default parameters ...

What's wrong there?

How do I forward default parameters correctly?

Answer 1

g may have default arguments, but the type of &g is still int(*)(int, int), which is not a type that can be called with no arguments. Within generic, we can't differentiate that - we've already lost the context about default arguments.

You can just wrap g in a lambda to preserve the context:

generic([]{ return g(); });