使用静态/全局const原语的计算是否在编译时进行？

Question

const int thing = 1234;

int func( int hello )
{
    return hello / (thing*123+321);
}

Is (thing*123+321) computed in compile time?

If thing was a user defined type with int operator* would it also happen in compile time?

If thing was a float or double would it also happen in compile time?

If I forcibly change thing in runtime would (thing*123+321) be unchanged?

Answer 1

There's no guarantee that it will be, but any optimizing compiler will precompute this value. It will probably turn that division operation into a multiplication by a precomputed reciprocal value too.

Answer 2

In general, constant expressions can and usually will be computed at compile time. Since you can and if you do use constant expressions as template arguments, there's actually no way around that.

The big question is now what constitutes a constant expression. In C++98/03, your thing would qualify as one, but a constant function of thing would not, since there's no way to tell the compiler that a function is "pure" in that sense.

Some compilers offer extensions (such as GCC's attributes "pure" and "const"), but that's outside the scope of the language. C++11 introduces the explicit constexpr keyword, which allows you to declare all sorts of expressions as constant and thus eligible for compile-time evaluation. (For instance, GCC is built with the GMP and MPFR libraries in order to perform those compile-time computations to any degree of precision, so as to be able to target any platform.)

Answer 3

Is (thing*123+321) computed in compile time?

Yes.

hello / (thing*123+321);

In this case, the full expression is computed at runtime. since the argument in your function hello is unknown at until you run the program.