c++ - 为什么 auto 的推论不同？

Question

int main(){
    int x{};
    auto x2 = x;
    auto x3{x};

    static_assert(is_same<int, decltype(x)>::value, "decltype(x) is the same as int");
    static_assert(is_same<int, decltype(x2)>::value, "decltype(x2) is the same as int");
    static_assert(is_same<int, decltype(x3)>::value, "decltype(x3) is the same as int"); // Error here.
}

此代码无法使用 gcc 4.8.0 编译。我什至不猜测decltype(x3). 它是什么？为什么行为不同？

Answer 1

#include <initializer_list>
#include <type_traits>

using namespace std;

int main(){
    int x{};
    auto x2 = x;
    auto x3{x};

    static_assert(is_same<int, decltype(x)>::value, "decltype(x) is the same as int");
    static_assert(is_same<int, decltype(x2)>::value, "decltype(x2) is the same as int");
    static_assert(is_same<std::initializer_list<int>, decltype(x3)>::value, "decltype(x3) is the same as int");
}

这将编译。x3被推断为std::initializer_list<int>由于：

让T是已为变量标识符确定的类型d。P如果初始值设定项是一个花括号初始化列表（8.5.4），则从 T [...]获取，使用std::initializer_list<U>.

Answer 2

所以x3实际上是一个std::initializer_list<int>，你想出这个的一种方法如下：

std::cout << typeid(x3).name() << std::endl ;

对我来说，我有以下输出：

St16initializer_listIiE

把它通过c++filt：

c++filt -t St16initializer_listIiE

给我：

std::initializer_list<int>