Why is the default value of a function 1?

Question

I noticed that the constructor with parentheses behaves differently, at least for floats and ints.

// int a, b{}, c = T(), d = T{}, e();
float a, b{}, c = T(), d = T{}, e();
endl(std::cout << a << ", " << b << ", " << c << ", " << d << ", " << e);

// 0, 0, 0, 0, 1

It's not a default constructor, the parentheses were parsed as part of the type. I printed the types, and e has type float (&)() (or just float() with proper forwarding), so I think it's a default-constructed function. The result of calling it has a type, but calling it results in the linker error you'd expect. Its value would be 0 if it were a function pointer. Why would its value be 1, or how does it end up as 1 when it's printed?

Answer 1

It's not a function pointer, it's literally a function (albeit an undefined one).

By streaming the expression e, which decays, you're creating a pointer to that function, and said pointer is valid because it points to the function you've declared called e. Hence, 1.

You're right to say that if you declared a function pointer to begin with, and made it nullptr, you'd see 0 instead by streaming it. But functions and function pointers are two different things.