Initializing a constexpr with a const, — int vs float

问题

I'm wondering why the integer ii is initiallized at compile time, but not the float ff here:

int main() {
  const int i = 1;
  constexpr int ii = i;

  const float f = 1.0;
  constexpr float ff = f;
 }

This is what happens when I try to compile:

> g++ -std=c++11 test.cc
test.cc: In function ‘int main()’:
test.cc:6:24: error: the value of ‘f’ is not usable in a constant expression
   constexpr float ff = f;
                        ^
test.cc:5:15: note: ‘f’ was not declared ‘constexpr’
   const float f = 1.0;

回答1:

Constant variables of integral types with constant initializers are integral constant expressions (de facto implicitely constexpr; see expr.const in ISO C++). float is not an integral type and does not meet the requirements for constant expression without the use of constexpr. (A similar case is why int can be but float cannot be a template parameter.)

回答2:

In C++ constant integers are treated differently than other constant types. If they are initialized with a compile-time constant expression they can be used in a compile time expression. This was done so that array size could be a const int instead of #defined (like you were forced in C):

(Assume no VLA extensions)

const int s = 10;
int a[s];          // OK in C++

来源：https://stackoverflow.com/questions/34665079/initializing-a-constexpr-with-a-const-int-vs-float