为什么 {} 作为函数参数不会导致歧义？

Question

考虑这段代码：

#include <vector>
#include <iostream>

enum class A
{
  X, Y
};

struct Test
{
  Test(const std::vector<double>&, const std::vector<int>& = {}, A = A::X)
  { std::cout << "vector overload" << std::endl; }

  Test(const std::vector<double>&, int, A = A::X)
  { std::cout << "int overload" << std::endl; }
};

int main()
{
  std::vector<double> v;
  Test t1(v);
  Test t2(v, {}, A::X);
}

https://godbolt.org/z/Gc_w8i

这会打印：

vector overload
int overload

为什么这不会由于不明确的重载解析而产生编译错误？如果删除第二个构造函数，我们会得到两次 vector overload。 How/by 什么指标 int 比 std::vector<int> 更明确地匹配 {}？

^{构造函数签名当然可以进一步修剪，但我只是被一段等效的代码欺骗了，想确保这个问题没有丢失任何重要内容。}

Answer 1

它在[over.ics.list]，强调我的

6 Otherwise, if the parameter is a non-aggregate class X and overload resolution per [over.match.list] chooses a single best constructor C of X to perform the initialization of an object of type X from the argument initializer list:

If C is not an initializer-list constructor and the initializer list has a single element of type cv U, where U is X or a class derived from X, the implicit conversion sequence has Exact Match rank if U is X, or Conversion rank if U is derived from X.

Otherwise, the implicit conversion sequence is a user-defined conversion sequence with the second standard conversion sequence an identity conversion.

9 Otherwise, if the parameter type is not a class:
[...]
if the initializer list has no elements, the implicit conversion sequence is the identity conversion. [ Example:
void f(int);
f( { } ); // OK: identity conversion
end example ]

std::vector 由构造函数初始化，粗体项目符号认为它是用户定义的转换。同时，对于 int，这是身份转换，因此它胜过第一个 c'tor 的等级。

为什么 {} 作为函数参数不会导致歧义？

Why does {} as function argument not lead to ambiguity?

c++

language-lawyer

overload-resolution