c ++模板函数参数推导和函数解析 [英] c++ template function argument deduce and function resolution

问题描述

今天我只想提出一个问题C ++模板函数参数演绎和模板函数重载解析在c ++ 11（我使用vs2010 sp1）。
我已经定义了两个模板函数如下：

Today I just want to raise a question on C++ template function argument deduce and template function overload resolution in c++ 11 (I am using vs2010 sp1). I have defined two template functions as below:

function＃1：

function #1:

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

function＃2：

function #2:

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

现在考虑下面的代码：

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);}
}

我只想知道引导函数重载解析在这些情况下。

I just want to know the rules behind guiding the function overloading resolution under these scenarios.

我不同意下面的两个答案。我认为const int示例不同于字符串示例。我可以修改#function 1一点，看看地球上推导出的类型

I don't agree with the two answers below.I think the const int example is different from the literal string example. I can modify the #function 1 a bit to see what’s the deduced type on earth

 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");

，const修饰符在const T&声明中， const int;而在文字字符串示例中，我不知道const修饰符在const T&声明中的位置。声明一些类似int& amp; const（但有意义的是声明int * const）

in the const int example, the const modifier in the "const T&" declaration eats up "the constness" of const int; while in the literal string example, I don’t know where the const modifier in the "const T&" declaration goes. It is meaningless to declare some like int& const (but it is meaningful to declare int* const)

推荐答案

这里的技巧是 / code>。 F1和F2都可以接受任何类型的任何值，但F2是一个更好的匹配，因为它是完美的转发。因此，除非值是 const lvalue，F2是最好的匹配。但是，当lvalue为 const 时，F1是更好的匹配。这就是为什么它是首选的const int和字符串文字。

The trick here is the const. Both F1 and F2 can accept any value of any type, but F2 is a better match in general, because it's perfect forwarding. So unless the value is a const lvalue, F2 is the best match. However, when the lvalue is const, F1 is the better match. This is why it's preferred for the const int and the string literal.

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