Why GCC does not evaluate constexpr at compile time?

问题

As an example:

class something {
public:
  static constexpr int seconds(int hour, int min, int sec)
  { return hour*3600+min*60+sec; }
}

then:

printf("Look at the time: %d\n", something::seconds(10, 0, 0));

Will compile to a call to the function using g++, instead of putting a constant number. Why would g++ do that? There's no gain in it and kinda defeats the purpose of using constexpr instead of awful macros.

回答1:

Why would g++ do that?

constexpr functions only must be evaluated at compile time in situations where the result is used as a constant expression. These include things like initializing a constexpr variable and being used as a template argument.

In other situations, even when a constexpr function is invoked with arguments that are all themselves constant expressions, it is up to the implementation to do what it wants. Typically, it'll depend on the optimization flags. On both gcc 6.2 and clang 3.9.1, for instance, -O0 will emit a call at runtime but -O1 will emit the constant 36000.

来源：https://stackoverflow.com/questions/41862118/why-gcc-does-not-evaluate-constexpr-at-compile-time