C++11：编译器何时将 {} 视为 std::initializer_list，何时不视为？

Question

我有一个快速示例：

#include <utility>

using namespace std;

struct A{
    int i;
    char c;
};

void f(const A&){}

template<class T>
void g(T&& t)
{
    f(forward<T>(t));
}

int main() {
    A a={1,'@'};//OK
    f({1,'#'});//OK
    g({1,'@'});//Compilation error
    return 0;
}

Clang 会给出这个错误：

    testArray.cpp:16:5: error: no matching function for call to 'g'
        g({1,'@'});//fix: g<A>({1,'@'})
        ^
    testArray.cpp:9:6: note: candidate template ignored: couldn't infer
          template argument 'T'
    void g(T&& t)
         ^

我的问题是：

在A a={1,'@'};中，如果{}推导为std::initializer_list，那么如何从std::initilizer_list转换为类型A ?
在f({1,'#'});中，当f需要类型A时，编译器是隐式生成一个A对象，还是从std::initializer_list 到 A?
为什么当g()是模板时，模板类型推导不能给出类型A？ std::forward 是否有助于将消息从 f 传达给 g，说 T 是类型 A？

Answer 1

In 'A a={1,'@'}'' :If {} is deduced as std::initializer_list, then how it's converted from std::initilizer_list to type A?

不，与std::initializer_list无关。 a 只是 copy-list-initialized 来自 {1,'@'}.

In 'f({1,'#'});' When f requires a type A, does compiler implicitly generates an A object, or it converts from std::initializer_list to A?

还是和std::initializer_list没有关系。函数参数是copy-list-initialized from {1,'#'}.

why when "g()" is a template, the template type deduction doesn't work to give a type A? Does "forward" help to convey the message from f to g, say that T is type A()?

因为这个属于non deduced context，模板参数T无法推导导致编译错误

6) The parameter P, whose A is a braced-init-list, but P is not std::initializer_list or a reference to one:

和

when does compiler consider {} as std::initializer_list, and when doesn't?

当您将函数参数类型声明为 std::initializer_list 并将花括号列表作为参数传递时，将构造 std::initializer_list。另一种情况是auto，例如

auto x1 = {3}; // x1 is deduced as std::initializer_list<int>
auto x2{1, 2}; // after C++17, error: not a single element
               // before C++17, it was std::initializer_list<int>
auto x3{3};    // after C++17, x3 is deduces as int
               // before C++17 it was std::initializer_list<int>

顺便说一句：即使使用 auto 它也不适用于 {1,'@'}，应该是 std::initializer_list<char> 还是 std::initializer_list<int>？

C++11：编译器何时将 {} 视为 std::initializer_list，何时不视为？

C++11: when does compiler consider {} as std::initializer_list, and when doesn't?

c++

templates

types

initializer-list

c++11