c++ - 为什么{}作为函数参数不会导致歧义？

Question

考虑以下代码:

#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

打印:

vector overload
int overload

为什么由于模棱两可的重载解析而不会产生编译错误？如果第二个构造函数被删除，我们将得到vector overload两次。 int如何/按哪种度量标准比{}更好地匹配std::vector<int>？

可以肯定地进一步减少构造函数的签名，但是我只是被等效的代码所欺骗，并希望确保对该问题没有任何重要的认识。

Answer 1

在 [over.ics.list] 中，重点是我的

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，粗体的项目符号将其视为用户定义的收敛。同时，对于int，这是身份转换，因此它胜过第一个c'tor的等级。