c++ - C++ 中的 constexpr int 和 constexpr double

Question

我遇到了一个奇怪的案例。

// this does not work, and reports:
// constexpr variable 'b' must be initialized by a constant expression
int main() {
  const double a = 1.0;
  constexpr double b = a;
  std::cout << b << std::endl;
}

// this works...
int main() {
  const int a = 1;
  constexpr int b = a;
  std::cout << b << std::endl;
}

double 有什么特别之处，因此它无法使 constexpr 工作吗？

Answer 1

anything special about double so it cannot make constexpr work?

int 和 double 在这种情况下表现不同。

对于core constant expression ，第二种情况下的 a(类型为 int)是 usable in constant expressions ，但第一种情况下的 a (类型为 double)则不是。

A core constant expression is any expression whose evaluation would not evaluate any one of the following:

• 9. an lvalue-to-rvalue implicit conversion unless....

     • a. applied to a non-volatile glvalue that designates an object that is usable in constant expressions (see below),

       int main() {
           const std::size_t tabsize = 50;
           int tab[tabsize]; // OK: tabsize is a constant expression
                             // because tabsize is usable in constant expressions
                             // because it has const-qualified integral type, and
                             // its initializer is a constant initializer
       
           std::size_t n = 50;
           const std::size_t sz = n;
           int tab2[sz]; // error: sz is not a constant expression
                         // because sz is not usable in constant expressions
                         // because its initializer was not a constant initializer
       }
       

和

(强调我的)

Usable in constant expressions In the list above, a variable is usable in constant expressions if it is

• a constexpr variable, or
• it is a constant-initialized variable
  • of reference type or
  • of const-qualified integral or enumeration type.

您可以将 a 声明为 constexpr 以使其可在常量表达式中使用。例如

constexpr double a = 1.0;
constexpr double b = a;   // works fine now