c++ - C++ POD 类型不能有任何构造函数吗？

Question

我有一个类和一个 const 变量。

struct A 
{
    int b;
};

A const a;

A类是POD，可以这样初始化。

A const a = { 3 };

恕我直言，拥有这样的构造函数看起来不错。

struct A 
{
    int b;

    A(int newB) : b(newB)
    {
    }
};

但 Clang 假定 A 为非聚合类型。为什么我不能有这样的构造函数？还是我应该做点别的？

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我修改了问题以表达我原来的意思。我错误地将 struct 写为 class，对于 @Johannes 的混淆感到抱歉:)

Answer 1

POD 表示普通旧数据类型，根据定义不能有用户定义的构造函数。

POD 实际上是一个聚合类型(见下一段引文)。那么什么是聚合？ C++ 标准在第 8.5.1/1 节中说，

An aggregate is an array or a class (clause 9) with no user-declared constructors (12.1), no private or protected nonstatic data members (clause 11), no base classes (clause 10), and no virtual functions (10.3).

C++ 标准的第 9/4 节说，

[....] A POD-struct is an aggregate class that has no non-static data members of type non-POD-struct, non-POD-union (or array of such types) or reference, and has no user-defined copy assignment operator and no user-defined destructor. Similarly, a POD-union is an aggregate union that has no non-static data members of type non-POD-struct, non-POD-union (or array of such types) or reference, and has no user-defined copy assignment operator and no user-defined destructor. A POD class is a class that is either a POD-struct or a POD-union.

由此可见，POD 类/结构/union 虽然不能有用户定义的赋值操作符和用户定义的析构函数。

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不过，还有其他类型的 POD。第 3.9/10 节说，

Arithmetic types (3.9.1), enumeration types, pointer types, and pointer to member types (3.9.2), and cv-qualified versions of these types (3.9.3) are collectively called scalar types. Scalar types, POD-struct types, POD-union types (clause 9), arrays of such types and cv-qualified versions of these types (3.9.3) are collectively called POD types.

阅读此常见问题解答:What is a "POD type"?