c++ 为什么Foo({})调用Foo(0)而不是Foo()？

可执行文件由clang 3.5.0和gcc 4.9.1从代码生成

#include <iostream>

struct Foo
{
   Foo() { std::cout << "Foo()" << std::endl; }
   Foo(int x) { std::cout << "Foo(int = " << x << ")" << std::endl; }
   Foo(int x, int y) { std::cout << "Foo(int = " << x << ", int = " << y << ")" << std::endl; }
};

int main()                 // Output
{                          // ---------------------
   auto a = Foo();         // Foo()
   auto b = Foo(1);        // Foo(int = 1)
   auto c = Foo(2, 3);     // Foo(int = 2, int = 3)
   auto d = Foo{};         // Foo()
   auto e = Foo{1};        // Foo(int = 1)
   auto f = Foo{2, 3};     // Foo(int = 2, int = 3)
   auto g = Foo({});       // Foo(int = 0)          <<< Why?
   auto h = Foo({1});      // Foo(int = 1)
   auto i = Foo({2, 3});   // Foo(int = 2, int = 3)
}

表现为评论.

从 cppreference: cpp/language/list initialization ：

06001

The effects of list initialization of an object of type T are:

If T is an aggregate type, aggregate initialization is performed.

Otherwise, If the braced-init-list is empty and T is a class type with a default constructor, value-initialization is performed.

06002

我得出结论,Foo({})应该调用默认构造函数.

这个bug在哪里？

默认构造函数仅适用于您使用一对大括号：

auto a = Foo();         // Foo()
auto b = Foo{};         // Foo()

Foo({})只会调用具有空列表的构造函数作为参数,copy-list-initialize任何构造函数的参数. [dcl.init] / 16：

If the destination type is a (possibly cv-qualified) class type:

— If

the initialization is direct-initialization […] constructors are considered. The applicable constructors

are enumerated (13.3.1.3), and the best one is chosen through overload

the overload resolution is ambiguous, the initialization is

ill-formed.

你有一个参数：空的braced-init-list.有一个列表初始化序列将{}转换为int,所以构造函数Foo(int)是通过重载分辨率来选择的.参数初始化为零,因为{}表示一个 value-intialization ,对于标量,这意味着一个 zero-initialization .

在cppreferences文档中也没有错误：对于(7)说明

7) in a functional cast expression or other direct-initialization,  with braced-init-list used as the constructor argument

这显然导致与上述引用相同的结果：使用(空)braced-init-list调用构造函数.