功能模板专业化和Abrahams / Dimov示例 [英] Function template specialization and the Abrahams/Dimov example

问题描述

（我假设您在此问题中了解 Abrahams / Dimov示例。 ）

(I'm assuming knowledge of the Abrahams/Dimov example in this question.)

假设在标题中有一些第三方代码，您无法修改：

Assume there is some 3rd-party code in a header that like this, which you cannot modify:

template<class T> void f(T);    // (1) base template 1
template<class T> void f(T *);  // (2) base template 2
template<> void f<>(int *);     // (3) specialization of (2)

问题是：

如果我已经给出了上面的声明，那么是可能的对于我现在专门化基本模板1的情况下 T = int * （例如）？

If I have been given the declarations above as-is, is it possible for me to now specialize the base template 1 for the case where T = int * (for example)?

或者仅仅声明基本模板2意味着基本模板1不能再被专门化至少对于指针）？

Or does the mere declaration of base template 2 imply that base template 1 can no longer be specialized (at least for pointers)?

推荐答案

您可以通过明确指定模板参数函数名（cf. C ++ 11-Standard 14.7.3）

You can overload (1) by explicitly specifying the template parameter in the angle-brackets after the function name (cf. C++11-Standard 14.7.3)

#include <iostream>
using namespace std;
template<class T> void f(T)    // (1) base template 1
{
    cout << "template<class T> void f(T)" << endl;
}

template<class T> void f(T *)  // (2) base template 2
{
    cout << "template<class T> void f(T *)" << endl;
}
//template<> void f<>(int *);     // (3) specialization of (2)

template<> void f<int*>(int *)     // (4) specialization of (1)
{
    cout << "f<int*>(int *)" << endl;
}


int main() {
    int i;
    f(&i); // calls (2) since only base-templates take part in overload resolution
    return 0;
}

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