c++模板函数参数推导和函数解析

Question

今天我只是想提出一个关于C++ 11(我使用的是vs2010 sp1)中的C++模板函数参数推导和模板函数重载解析的问题。 我定义了两个模板函数如下:

函数#1:

template <class T>
void func(const T& arg)
{
    cout << "void func(const T&)" <<endl;
}

函数#2:

template <class T>
void func(T&& arg)
{
   cout << "void func(T&&)" <<endl;
}

现在考虑以下代码:

int main() {
    //I understand these first two examples:

    //function #2 is selected, with T deduced as int&
    //If I comment out function #2, function#1 is selected with
    //T deduced as int
    {int a = 0; func(a);}

    //function #1 is selected, with T is deduced as int.
    //If I comment out function #1, function #2 is selected,
    //with T deduced as const int&.
    {const int a = 0; func(a);}

    //I don't understand the following examples:  

    //Function #2 is selected... why?
    //Why not function #1 or ambiguous...
    {func(0);}

    //But here function #1 is selected.
    //I know the literal string “feng” is lvalue expression and
    //T is deduced as “const char[5]”. The const modifier is part
    //of the T type not the const modifier in “const T&” declaration. 
    {func(“feng”)}

    //Here function#2 is selected in which T is deduced as char(&)[5]
    {char array[] = “feng”; func(array);}
}

我只想知道在这些场景下指导函数重载决策背后的规则。

我不同意下面的两个答案。我认为 const int 示例与文字字符串示例不同。我可以稍微修改一下#function 1 看看推导的类型到底是什么

 template <class T>
 void func(const T& arg)
 {
    T local;
    local = 0;
    cout << "void func(const T&)" <<endl;
 }
 //the compiler compiles the code happily 
 //and it justify that the T is deduced as int type
 const int a = 0;
 func(a);

 template <class T>
 void func(const T& arg)
 {
T local;
Local[0] = ‘a’;
cout << "void func(const T&)" <<endl;
 }
 //The compiler complains that “error C2734: 'local' : const object must be     
 //initialized if not extern
 //see reference to function template instantiation 
 //'void func<const char[5]>(T (&))' being compiled
  //    with
  //    [
  //        T=const char [5]
  //    ]

 Func(“feng”);

在 const int 示例中，“const T&”声明中的 const 修饰符吞噬了 const int 的“constness”；而在文字字符串示例中，我不知道“const T&”声明中的 const 修饰符在哪里。声明一些像int& const是没有意义的(但是声明int* const是有意义的)

Answer 1

这里的技巧是 const。 F1 和 F2 都可以接受任何类型的任何值，但 F2 通常是更好的匹配，因为它是完美转发。因此除非值是 const 左值，否则 F2 是最佳匹配。然而，当左值为 const 时，F1 是更好的匹配。这就是为什么它更适合 const int 和字符串文字的原因。