为什么模板参数扣除不工作在这里？ [英] Why is the template argument deduction not working here?

问题描述

我创建了两个简单的函数，它们获得模板参数和一个定义类型的空结构：

I created two simple functions which get template parameters and an empty struct defining a type:

//S<T>::type results in T&
template <class T>
struct S
{
    typedef typename T& type;
};

//Example 1: get one parameter by reference and return it by value
template <class A>
A
temp(typename S<A>::type a1)
{
    return a1;
}

//Example 2: get two parameters by reference, perform the sum and return it
template <class A, class B>
B
temp2(typename S<A>::type a1, B a2)//typename struct S<B>::type a2)
{
    return a1 + a2;
}

参数类型应用于struct S以获取引用。我用一些整数值调用它们，但编译器无法推导出参数：

The argument type is applied to the struct S to get the reference. I call them with some integer values but the compiler is unable to deduce the arguments:

int main()
{
    char c=6;
    int d=7;
    int res = temp(c);
    int res2 = temp2(d,7);
}

错误1错误C2783：'A
temp（S :: type）'：无法推导出'A'的
模板参数

Error 1 error C2783: 'A temp(S::type)' : could not deduce template argument for 'A'

错误2错误C2783：'B
temp2（S :: type，B）'：can not
推导'A'的模板参数

Error 2 error C2783: 'B temp2(S::type,B)' : could not deduce template argument for 'A'

---

为什么会发生这种情况？难道看到模板参数是 char 和 int 值吗？

推荐答案

正如第一个注意，当你提到一个从属名称时，会使用typename名称。所以你不需要它。

Just as first note, typename name is used when you mention a dependent name. So you don't need it here.

template <class T>
struct S
{
    typedef T& type;
};

< ：type 表示A的非隐藏上下文。当模板参数仅用于非隐藏上下文（在函数中为A的情况）时，模板不被考虑参数扣除。详细信息在C ++标准（2003）的第14.8.2.4节。

Regarding the template instantiation, the problem is that typename S<A>::type characterizes a nondeduced context for A. When a template parameter is used only in a nondeduced context (the case for A in your functions) it's not taken into consideration for template argument deduction. The details are at section 14.8.2.4 of the C++ Standard (2003).

要使调用工作，您需要明确指定类型：

To make your call work, you need to explicitly specify the type:

temp<char>(c);

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