部分模板专业化歧义 [英] Partial template specialization ambiguity

问题描述

我不明白为什么main中的语句不明确.

I cant see why the statement in main is ambiguous.

template<class T, class U, int I> struct X
{ void f() { cout << "Primary template" << endl; } };


template<class T, int I> struct X<T, T*, I>
{void f() { cout << "Partial specialization 1" << endl;}};

template<class T, class U, int I> struct X<T*, U, I>
{void f() { cout << "Partial specialization 2" << endl;}};

template<class T> struct X<int, T*, 10>
{void f() { cout << "Partial specialization 3" << endl;}};

template<class T, class U, int I> struct X<T, U*, I>
{void f() { cout << "Partial specialization 4" << endl;}};

 int main()
 {
   X<int, int*, 10> f;
 }

不是最专业的模板吗? 这是来自

Isn't X<int, T*, 10> the most specialized template? This is an example from http://publib.boulder.ibm.com/infocenter/lnxpcomp/v8v101/index.jsp?topic=%2Fcom.ibm.xlcpp8l.doc%2Flanguage%2Fref%2Fpartial_specialization.htm

推荐答案

如果每个与第一个匹配的参数列表也与第二个匹配，而与之相反，则模板专用化比另一个专用化.

A template specialization is more specialized than another if every argument list that matches the first also matches the second, but not the other way around.

查看X<int, T*, 10>和X<T, T*, I>时:

• X<int, float*, 10>与第一个匹配，但与第二个不匹配.
• X<float, float*, 10>与第二个匹配，但与第一个不匹配.

• X<int, float*, 10> matches the first but not the second.
• X<float, float*, 10> matches the second but not the first.

因此，任何一个都不比另一个更专业，并且匹配这两个专业的模板实例化也不会编译.

Therefore neither is more specialized than the other, and a template instantiation that matches both specializations won't compile.

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