constexpr void函数被拒绝 [英] constexpr void function rejected

问题描述

我有一个非常简单的函数，不会编译.

I have this very simple function which won't compile.

constexpr void func()
{
}

我得到的错误是:

error: invalid return type 'void' of constexpr function 'constexpr void func()'

     constexpr void func()

在C ++ 14中，void是文字类型[§3.9/10]:

In C++14, void is a literal type [§3.9/10]:

如果是，则类型是文字类型:

• void;或
• 标量类型；或
• 参考类型；或
• 文字类型的数组；或
• 具有以下所有属性的类类型(第9条):
  • 它有一个琐碎的析构函数，
  • 它是聚合类型(8.5.1)或具有至少一个不是复制或移动构造函数的constexpr构造函数或构造函数模板，并且
  • 其所有非静态数据成员和基类均为非易失性文字类型.
  • void; or
  • a scalar type; or
  • a reference type; or
  • an array of literal type; or
  • a class type (Clause 9) that has all of the following properties:
    • it has a trivial destructor,
    • it is an aggregate type (8.5.1) or has at least one constexpr constructor or constructor template that is not a copy or move constructor, and
    • all of its non-static data members and base classes are of non-volatile literal types.

    有人可以解释为什么这无效吗?

    Can someone explain why this is invalid?

    推荐答案

    使void成为文字类型的提案是 n3652对constexpr函数的放宽约束 . G ++决定将此功能推至版本 5 (我使用的是4.9.2):

    The proposal which made void a literal type was n3652 Relaxing constraints on constexpr functions. G++ decided to push this feature to version 5 (I was using 4.9.2):

    G ++现在支持C ++ 14扩展的constexpr.

    G++ now supports C++14 extended constexpr.

    constexpr int f (int i)
    {
      int j = 0;
      for (; i > 0; --i)
        ++j;
      return j;
    }
    
    constexpr int i = f(42); // i is 42
    

    从3.4版开始，Clang就实现了这一点.

    Clang has had this implemented since version 3.4.

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