大括号的数量如何影响统一初始化?

问题描述

请考虑以下代码段:

#include <iostream>

struct A {
  A() {}
  A(const A&) {}
};

struct B {
  B(const A&) {}
};

void f(const A&) { std::cout << "A" << std::endl; }
void f(const B&) { std::cout << "B" << std::endl; }

int main() {
  A a;
  f(   {a}   ); // A
  f(  {{a}}  ); // ambiguous
  f( {{{a}}} ); // B
  f({{{{a}}}}); // no matching function
}

为什么每个调用都构造相应的输出?大括号的数量如何影响统一初始化?括号删除如何影响所有这些?

Why does each call fabricate the corresponding output? How does the number of braces affect uniform initialization? And how does brace elision affect all this?

推荐答案

过载分辨率很有趣.

1. {a}具有完全匹配等级，用于初始化(临时)const A&参数，该参数胜过用户定义的转换B(const A&)作为{a}的实现. 在C ++ 14中添加了此规则，以解决列表初始化中的歧义(以及对聚合的调整).

1. {a} has exact match rank for initializing (a temporary for) the const A& parameter, which outcompetes the user-defined conversion B(const A&) as a realization of {a}. This rule was added in C++14 to resolve ambiguities in list-initialization (along with adjustments for aggregates).

请注意，名义临时名称是从未创建:在重载解析选择f(const A&)之后，引用只是初始化为引用a ，并且此解释甚至可以应用于不可复制的类型.

Note that the notional temporary is never created: after overload resolution picks f(const A&), the reference is simply initialized to refer to a, and this interpretation can apply even for non-copyable types.

不涉及括号删除-我们不知道允许使用的最外面的目标类型.

No brace elision is involved—we don’t know the outermost target type to allow it.

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