c++ - 在常量初始值设定项中调用 constexpr 构造函数的限制

Question

我不明白为 constexpr constructor 调用添加的以下内容:

A constant initializer for an object o is an expression that is a constant expression, except that it may also invoke constexpr constructors for o and its subobjects even if those objects are of non-literal class types

Constexpr 构造函数是核心常量表达式本身:

A conditional-expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine (1.9), would evaluate one of the following expressions:

[...]

— an invocation of a function other than a constexpr constructor for a literal class, a constexpr function, or an implicit invocation of a trivial destructor (12.4)[...]

Answer 1

后面的注释中有实质性提示:

[ Note: such a class may have a non-trivial destructor — end note ]

回想一下文字类类型 (3.9p10):

• has a trivial destructor,
  
• 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.

因此，如果一个类有一个非平凡的析构函数，它就没有资格成为文字类，但对其 constexpr 构造函数的调用可能仍然有资格作为常量初始化程序:

#include <iostream>
struct A {
    int count = 0;
    constexpr A() {}
    ~A() { std::cout << "~A: " << count << std::endl; }
};
A a;

然后程序可以依赖 a 在任何动态初始化发生之前被初始化，即使 A 不是文字类类型。

这样做的理由是虽然A的析构函数有副作用，但在初始化期间不会调用它，因此编译器可以计算出适当的编译时 a 的初始内存内容(实际上，它的 .data)。

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请注意，您在第二个引用中的强调是不完整的；相关术语是“a constexpr constructor for a literal class”。