初始化constexpr静态类成员时的编译器错误 [英] Compiler error when initializing constexpr static class member

问题描述

我已经通过以下方式声明了一个类

I've declared a class in the following way

class A
{
    struct B
    {
        constexpr
        B(uint8_t _a, uint8_t _b) :
            a(_a),
            b(_b)
        {}

        bool operator==(const B& rhs) const
        {
            if((a == rhs.a)&&
               (b == rhs.b))
            {
                return true;
            }
            return false;
        }

        uint8_t a;
        uint8_t b;
    };

    constexpr static B b {B(0x00, 0x00)};

};

但是g ++表示

错误：字段初始值设定项不是常量

error: field initializer is not constant

无法弄清楚我错了。

推荐答案

C语更有用：

27 : error: constexpr variable 'b' must be initialized by a constant expression
constexpr static B b {B(0x00, 0x00)};
                   ^~~~~~~~~~~~~~~~
27 : note: undefined constructor 'B' cannot be used in a constant expression
constexpr static B b {B(0x00, 0x00)};
                      ^
8 : note: declared here
B(uint8_t _a, uint8_t _b) :
^

在成员变量的
brace-or-equal-initializer 中，构造函数（包括嵌套类的构造函数）被认为是未定义的；这是因为对于构造函数而言，引用成员变量的值是合法的，因此即使在词法上它们后来位于文件中，也必须首先定义成员变量：

Within a brace-or-equal-initializer of a member variable, constructors (including constructors of nested classes) are considered undefined; this is because it is legitimate for a constructor to refer to the values of member variables, so the member variables must be defined first even if they are lexically later in the file:

struct A {
  struct B { int i; constexpr B(): i{j} {} };
  constexpr static int j = 99;
};

解决方法是将 B 放在 A ，或者在基类中。

The workaround is to place B outside A, or perhaps within a base class.

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