Why can function pointers be `constexpr`?

Question

How does the compiler know where in memory the square root will be before the program is executed? I thought the address would be different everytime the program is executed, but this works:

constexpr double(*fp)(double) = &sqrt; cout << fp(5.0); 

Is it because the address is relative to another address in memory? I don't think so because the value of fp is large: 0x720E1B94.

Answer 1

At compile time, the compiler doesn't know the address of sqrt. However, you cannot do anything at compile time with a constexpr function pointer that would allow you to access that pointer's address. Therefore, a function pointer at compile time can be treated as an opaque value.

And since you can't change a constexpr variable after it has been initialized, every constexpr function pointer can be boiled down to the location of a specific function.

If you did something like this:

using fptr = float(*)(float);  constexpr fptr get_func(int x) {   return x == 3 ? &sqrtf : &sinf; }  constexpr fptr ptr = get_func(12); 

The compiler can detect exactly which function get_func will return for any particular compile time value. So get_func(12) reduces down to &sinf. So whatever &sinf would compile to is exactly what get_func(12) would compile to.