c ++模板函数参数推导和函数解析

Question

今天我只想提出一个关于C++模板函数参数推导和模板函数重载决议的问题在c ++ 11中(我使用的是vs2010 sp1).我已经定义了两个模板函数,如下所示:

功能#1:

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

功能#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引导功能重载解决方案背后的规则.

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#function 1以查看地球上推断出的类型

 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修饰符吃掉了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和const是没有意义的(但声明int*const是有意义的)

Answer 1

这里的诀窍是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,因为它是完美的转发.因此,除非该值是const左值,否则F2是最佳匹配.但是,当左值是const,F1是更好的匹配.这就是为什么它首选const int和字符串文字.