c++ - 模板类型推导执行隐式数组到指针的转换

Question

我已阅读有关此标准转换的现有问题。但是，我没有找到满意的答案。

我有这段代码显示选择了 T* 重载而不是 T&& 重载。据我了解，转发引用重载应该绑定(bind)所有内容，除非另一个重载是完美匹配。

在下面的代码中，tab 是一个char const(&)[4]。

有人能给我解释一下为什么要在这里进行数组到指针的转换吗？如果有办法解决该转换，我会洗耳恭听!

(大肠杆菌 link )

#include <utility>

template <typename T>
void f(T&& lol)
{
}

template <typename T>
void f(T* pof)
{
  static_assert(sizeof(T) && false, "");
}

template <typename T>
struct S;

int main(int argc, char *argv[])
{
  decltype("lol") tab = "lol";
  S<decltype(tab)> s;
  f("lol");
  return 0;
}

Answer 1

数组被推导为 Ptr 类型。

参见 http://en.cppreference.com/w/cpp/language/template_argument_deduction

Before deduction begins, the following adjustments to P and A are made:

1) If P is not a reference type,

a) if A is an array type, A is replaced by the pointer type obtained from array-to-pointer conversion;

b) otherwise, if A is a function type, A is replaced by the pointer type obtained from function-to-pointer conversion;

c)otherwise, if A is a cv-qualified type, the top-level cv-qualifiers are ignored for deduction