c++ - 对 C++ 的哪些更改使复制初始化适用于具有显式构造函数的类？

Question

考虑这段代码:

struct X{
    explicit X(){}
    explicit X(const X&){}
};

void foo(X a = X()){}

int main(){}

使用 C++14 标准，GCC 7.1 和 clang 4.0 rejects代码，这是我所期望的。

但是，使用 C++17 (-std=c++1z)，they both accept编码。改变了什么规则？

---

对于两个编译器都表现出相同的行为，我怀疑这是一个错误。但据我所知，最新的草案仍然说，默认参数使用 copy-initialization ¹ 的语义。同样，我们知道 explicit 构造函数将只允许 直接初始化 ²。

¹:dcl.fct.default/5 ; ²: class.conv.ctor/2

Answer 1

因为 copy elision 的行为从 C++17 更改；在这种情况下，复制省略是强制性的。

Mandatory elision of copy/move operations

Under the following circumstances, the compilers are required to omit the copy and move construction of class objects, even if the copy/move constructor and the destructor have observable side-effects. The objects are constructed directly into the storage where they would otherwise be copied/moved to. The copy/move constructors need not be present or accessible:

• In the initialization of an object, when the initializer expression is a prvalue of the same class type (ignoring cv-qualification) as the variable type:

  T f() {
      return T();
  }
  
  T x = T(T(f())); // only one call to default constructor of T, to initialize x
  

Note: the rule above does not specify an optimization: C++17 core language specification of prvalues and temporaries is fundamentally different from that of the earlier C++ revisions: there is no longer a temporary to copy/move from. Another way to describe C++17 mechanics is "unmaterialized value passing": prvalues are returned and used without ever materializing a temporary.

对于 copy initialization :

The effects of copy initialization are:

• First, if T is a class type and the initializer is a prvalue expression whose cv-unqualified type is the same class as T, the initializer expression itself, rather that a temporary materialized from it, is used to initialize the destination object: see copy elision (since C++17)

• If T is a class type and the cv-unqualified version of the type of other is T or a class derived from T, the non-explicit constructors of T are examined and the best match is selected by overload resolution. The constructor is then called to initialize the object.

这意味着对于 X a = X()，a 将默认直接构造，复制/移动构造函数及其副作用将被完全省略。不会为重载决议选择非显式构造函数，这在 C++14(及之前)中是必需的。对于这些有保证的情况，复制/移动构造函数不参与，那么它们是否是 explicit 无关紧要。