blocks|key|5111399|text|函数模板不能部分专门化。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5111400|另一种方法是将其定义为静态成员函数(或者在本例中只定义一个值！)类模板，然后只提供一个普通函数模板包装器，如…|5111401|#include+<type_traits>

template<+class+T1,+class+T2,+class+T3+>
struct+Blah+{};

namespace+detail+{
++++template<+class+T1,+class+T2,+class+T3+>
++++struct+IsHomogenous+{+static+bool+const+yes+=+false;+};

++++template<+class+T+>
++++struct+IsHomogenous<+T,+T,+T+>+{+static+bool+const+yes+=+true;+};
}++//+namespace+detail

template<+class+T1,+class+T2,+class+T3+>
bool+isHomogenous(+Blah<+T1,+T2,+T3+>+)
{
++++return+detail::IsHomogenous<+T1,+T2,+T3+>::yes;
}

#include+<iostream>
int+main()
{
++++using+namespace+std;
++++wcout+<<+boolalpha
++++++++<<+isHomogenous(+Blah<+double,+char,+void+>()+)+<<+"+"
++++++++<<+isHomogenous(+Blah<+int,+int,+int+>()+)
++++++++<<+endl;
}|code-block|syntax|javascript|5111402|另一种方法是使用C%2B%2B11+std::is_same|offset|length|style|CODE|5111403|#include+<type_traits>

template<+class+T1,+class+T2,+class+T3+>
struct+Blah+{};

template<+class+T1,+class+T2,+class+T3+>
bool+isHomogenous(+Blah<+T1,+T2,+T3+>+)
{
++++using+std::is_same;
++++return+is_same<+T1,+T2+>::value+&&+is_same<+T2,+T3+>::value;
}

#include+<iostream>
int+main()
{
++++using+namespace+std;
++++wcout+<<+boolalpha
++++++++<<+isHomogenous(+Blah<+double,+char,+void+>()+)+<<+"+"
++++++++<<+isHomogenous(+Blah<+int,+int,+int+>()+)
++++++++<<+endl;
}|5111404|entityMap^0|0|0|0|E|C|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|V|8|@$K|W|L|X|M|N]]|9|@]|A|$]]|$1|O|3|P|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

A function template cannot be partially specialized.

One alternative is to define it as a static member function (or in your case just a value!) of a class template, and then just provide an ordinary function template wrapper, like &hellip;

<pre><code>#include &lt;type_traits&gt;

template&lt; class T1, class T2, class T3 &gt;
struct Blah {};

namespace detail {
 template&lt; class T1, class T2, class T3 &gt;
 struct IsHomogenous { static bool const yes = false; };

 template&lt; class T &gt;
 struct IsHomogenous&lt; T, T, T &gt; { static bool const yes = true; };
} // namespace detail

template&lt; class T1, class T2, class T3 &gt;
bool isHomogenous( Blah&lt; T1, T2, T3 &gt; )
{
 return detail::IsHomogenous&lt; T1, T2, T3 &gt;::yes;
}

#include &lt;iostream&gt;
int main()
{
 using namespace std;
 wcout &lt;&lt; boolalpha
 &lt;&lt; isHomogenous( Blah&lt; double, char, void &gt;() ) &lt;&lt; " "
 &lt;&lt; isHomogenous( Blah&lt; int, int, int &gt;() )
 &lt;&lt; endl;
}
</code></pre>

and another way is to use C++11 <code>std::is_same</code>:

<pre><code>#include &lt;type_traits&gt;

template&lt; class T1, class T2, class T3 &gt;
struct Blah {};

template&lt; class T1, class T2, class T3 &gt;
bool isHomogenous( Blah&lt; T1, T2, T3 &gt; )
{
 using std::is_same;
 return is_same&lt; T1, T2 &gt;::value &amp;&amp; is_same&lt; T2, T3 &gt;::value;
}

#include &lt;iostream&gt;
int main()
{
 using namespace std;
 wcout &lt;&lt; boolalpha
 &lt;&lt; isHomogenous( Blah&lt; double, char, void &gt;() ) &lt;&lt; " "
 &lt;&lt; isHomogenous( Blah&lt; int, int, int &gt;() )
 &lt;&lt; endl;
}
</code></pre>

blocks|key|5021416|text|您不能专门化模板函数，当您添加专门化时，专门化函数实际上被添加到重载集中，因此它与重载相同，并使用重载解析规则来选择调用哪个函数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5021417|如果你真的想专门化一个模板函数，委托给一个具有静态成员的结构或类，然后专门化模板结构：|5021418|template+<typename+T>
struct+handler
{
+++static+void+perform(T&+);
};

template+<>
struct+handler<int>
{
+++static+void+perform(int)+{+doStuff();+}
};|code-block|syntax|javascript|5021419|下面是全局函数(您希望对其进行专门化)：|5021420|template+<typename+T>
void+function+(T&+v)
{
+++handler<T>::perform(v);
}|5021421|希望这能有所帮助。|5021422|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|U|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

You cannot specialize a template function, when you add a specialization the specialized function is actually added to the overload set so it is the same as overloading, and for choosing which function to call the overload resolution rules are used.

If you really want to specialize a template function, delegate to a struct or class with a static member and specialize the template struct instead:

<pre><code>template &lt;typename T&gt;
struct handler
{
 static void perform(T&amp; );
};

template &lt;&gt;
struct handler&lt;int&gt;
{
 static void perform(int) { doStuff(); }
};
</code></pre>

And here is the global function (which you want to specialize):

<pre><code>template &lt;typename T&gt;
void function (T&amp; v)
{
 handler&lt;T&gt;::perform(v);
}
</code></pre>

Hope this helps.

blocks|key|3380777|text|您可以使用函数部分专门化一个类，该函数将返回true或false。(或静态成员，等等)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3380778|template<+class+T1,+class+T2,+class+T3+>
class+Blah+{};

template<+class+T>
struct+isHomogenous
{
++++const+bool+operator()()+{+return+false;+}
};

template<+template+<typename,typename,typename>+class+T,+class+T1,+class+T2,+class+T3+>
struct+isHomogenous<T<T1,T2,T3>>
{
++++const+bool+operator()()+{+return+false;+}
};



template<+template+<typename,typename,typename>+class+T,+class+T1>
struct+isHomogenous<T<T1,T1,T1>>
{
++++const+bool+operator()()+{+return+true;+}
};

std::cout+<<+isHomogenous<+Blah<double,+char,+void>+>()()+<<+"+"
++++<<+isHomogenous<+Blah<int,+int,+int>+>()();|code-block|syntax|javascript|3380779|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You could partially specialize a class with a functor that will return the true or false. (or a static member, whatever):

<pre><code>template&lt; class T1, class T2, class T3 &gt;
class Blah {};

template&lt; class T&gt;
struct isHomogenous
{
 const bool operator()() { return false; }
};

template&lt; template &lt;typename,typename,typename&gt; class T, class T1, class T2, class T3 &gt;
struct isHomogenous&lt;T&lt;T1,T2,T3&gt;&gt;
{
 const bool operator()() { return false; }
};



template&lt; template &lt;typename,typename,typename&gt; class T, class T1&gt;
struct isHomogenous&lt;T&lt;T1,T1,T1&gt;&gt;
{
 const bool operator()() { return true; }
};

std::cout &lt;&lt; isHomogenous&lt; Blah&lt;double, char, void&gt; &gt;()() &lt;&lt; " "
 &lt;&lt; isHomogenous&lt; Blah&lt;int, int, int&gt; &gt;()();
</code></pre>

I have to implement a non-member function isHomogenous(Triple triple) for a template class defined as:

<pre><code>template &lt;typename T1, typename T2, typename T3&gt;
 class Triple
{
public:
 Triple()
 { }
 Triple(const T1 &amp;a, const T2 &amp;b, const T3 &amp;c) : a(a), b(b), c(c)
 { }
...
</code></pre>

The isHomogenous function should return a bool value indicating whether all of the three values in the parameter triple are of the same type. I have tried:

<pre><code>template &lt;typename T&gt; bool isHomogenous(Triple&lt;T, T, T&gt; triple) {
 return true;
}

template &lt;typename T1, typename T2, typename T3&gt; bool isHomogenous(Triple&lt;T1, T2, T3&gt; triple) {
 return false;
}
</code></pre>

This doesn't work, can you hint me on a solution?

C++ template function specialization

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我必须为一个模板类实现一个非成员函数isHomogenous(三元组)，其定义如下：template <typename T1, typename T2, typename T3> class Triple{public: Triple() { } Triple(const T1 &a, con...

问C++模板函数专门化
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