问题描述
我创建了两个简单的函数,它们获取模板参数和一个定义类型的空结构:
//S::type 结果为 T&模板结构体{typedef typename T&类型;};//示例1:通过引用获取一个参数并通过值返回模板<A类>一个temp(typename S<A>::type a1){返回 a1;}//例2:通过引用获取两个参数,求和返回模板<A类,B类>乙temp2(typename S<A>::type a1, B a2)//typename struct S<B>::type a2){返回 a1 + a2;}
参数类型应用于结构体 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:
int main(){字符 c=6;整数 d=7;int res = temp(c);int res2 = temp2(d,7);}
错误 1 错误 C2783:'Atemp(S::type)' : 无法推断'A' 的模板参数
错误 2 错误 C2783:'Btemp2(S::type,B)' : 不能推导出 'A' 的模板参数
为什么会这样?很难看出模板参数是 char 和 int 值吗?
正如第一个注意事项,当您提到一个从属名称时,会使用 typename 名称.所以你在这里不需要它.
模板 结构体{typedef T& type;};
关于模板实例化,问题在于 typename S<A>::type
描述了 A 的 nondeduced 上下文.当模板参数仅用于非推导的上下文(函数中 A 的情况)不考虑模板参数推导.详细信息位于 C++ 标准 (2003) 的第 14.8.2.4 节.
要使您的通话正常工作,您需要明确指定类型:
温度(c);
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;
}
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);
}
Why is this happening? Is it that hard to see that the template arguments are char and int values?
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;
};
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).
To make your call work, you need to explicitly specify the type:
temp<char>(c);
这篇关于为什么模板参数推导在这里不起作用?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!