直接初始化中的转换运算符

Question

C++14 标准 (N4296) 在 8.5/17.6.1

中说

If the initialization is direct-initialization [...], constructors are considered. The applicable constructors are enumerated, and the best one is chosen through overload resolution. [...] If no constructor applies, or the overload resolution is ambiguous, the initialization is ill-formed.

因此在直接初始化中，只考虑构造函数——忽略转换函数。在下面的代码中没有 A 的适用构造函数，只有 B 的转换函数。但是，代码编译通过，为什么？

struct A{};
struct B{
    operator A(){ return A{}; }
};

int main() {
    B b;
    A a(b);  // direct-initialization
}

Answer 1

您是正确的，在执行 A a(b); 时只考虑 A 的构造函数。 [over.match.ctor]/1 个州

When objects of class type are direct-initialized, copy-initialized from an expression of the same or a derived class type ([dcl.init]), or default-initialized, overload resolution selects the constructor. For direct-initialization or default-initialization that is not in the context of copy-initialization, the candidate functions are all the constructors of the class of the object being initialized. For copy-initialization (including default initialization in the context of copy-initialization), the candidate functions are all the converting constructors ([class.conv.ctor]) of that class. The argument list is the expression-list or assignment-expression of the initializer.

^强调我的

这意味着A()、A(const A&)和A(A&&)是候选列表。然后我们有 [over.match.viable]/4

[...]Third, for F to be a viable function, there shall exist for each argument an implicit conversion sequence that converts that argument to the corresponding parameter of F.[..]

允许将 b 隐式转换为 A，以便可以调用 A(A&&)。