blocks|key|4143903|text|在第二种情况下，getA不再是一个函数，而是一个函数模板，您不能有指向函数模板的指针。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4143904|你能做的就是让pfunction指向一个特定的getA实例(例如:对于T+=+int)：|4143905|class+MyClass
{
++++static+double+(*pfunction)(const+Object<int>+*,+const+Object<int>+*);
};

double+(*MyClass::pfunction)(const+Object<int>+*o1,+const+Object<int>+*o2)++=+&SomeClass::getA<int>;|code-block|syntax|javascript|4143906|但我不认为有一种方法可以让pfunction指向任何可能的getA实例。|4143907|entityMap^0|8|4|0|7|9|N|4|Z|7|0|0|D|9|T|4|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|P|$]]

In the second case, <code>getA</code> is not a function anymore but a function template, and you can't have a pointer to function template.

What you can do is have <code>pfunction</code> point to a particular <code>getA</code> instance (ie: for <code>T = int</code>) :

<pre><code>class MyClass
{
 static double (*pfunction)(const Object&lt;int&gt; *, const Object&lt;int&gt; *);
};

double (*MyClass::pfunction)(const Object&lt;int&gt; *o1, const Object&lt;int&gt; *o2) = &amp;SomeClass::getA&lt;int&gt;;
</code></pre>

But I don't think there is a way to get <code>pfunction</code> to point on any possible instance of <code>getA</code>.

blocks|key|1637020|text|template是一个模板+:)它不是一个具体的类型，不能用作成员。例如，您不能定义以下类：|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|1637021|class+A
{
++++template+<class+T>+std::vector<T>+member;
}|code-block|syntax|javascript|1637022|因为template+<class+T>+std::vector<T>+member;是潜在地可以专用于许多不同类型的东西。你可以这样做：|CODE|1637023|template+<class+T>
struct+A
{
+static+T+(*pfunction)();
};

struct+B
{
+template+<class+T>
+static+T+getT();
};

int+(*A<int>::pfunction)()+=+&B::getT<int>;|1637024|这里的A<int>是一个专门的模板，因此也有专门的成员|1637025|entityMap^0|B|2|0|0|2|15|0|0|3|6|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|W|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|X|8|@$9|Y|A|Z|B|M]]|D|@]|E|$]]|$1|N|3|O|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|P|3|Q|5|6|7|11|8|@$9|12|A|13|B|M]]|D|@]|E|$]]|$1|R|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|S|$]]

template is a template :) it's not a concrete type and cannot be used as a member. e.g. you cannot define following class:

<pre><code>class A
{
 template &lt;class T&gt; std::vector&lt;T&gt; member;
}
</code></pre>

because <code>template &lt;class T&gt; std::vector&lt;T&gt; member;</code> is something that potentially can be specialized to many different types. you can do something like this:

<pre><code>template &lt;class T&gt;
struct A
{
 static T (*pfunction)();
};

struct B
{
 template &lt;class T&gt;
 static T getT();
};

int (*A&lt;int&gt;::pfunction)() = &amp;B::getT&lt;int&gt;;
</code></pre>

here <code>A&lt;int&gt;</code> is a specialized template and so has specialized member

blocks|key|1633768|text|template+<class+T>
static+T+(+*pfunction+)+(+const+Object+<T>+*,+const+Object+<T>+*);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1633769|函数指针的模板在C%2B%2B中非法。它可以在类内，也可以简单地在类外。您不能这样写(甚至不能在类之外)：|unstyled|1633770|template+<class+X>
void+(*PtrToFunction)+(X);|1633771|请参阅此示例：http://www.ideone.com/smh73|offset|length|1633772|C%2B%2B标准说是$14/1，|1633773|1633774|模板定义了一系列类或函数。|blockquote|style|BOLD|1633775|1633776|1633777|请注意，它并没有说"A+template+defines+a+family+of+classes，function+or+function+pointers“。所以你要做的是，使用模板定义“一系列函数指针”，这是不允许的。|1633778|Loki库中的泛型函数库可以很好地解决您遇到的这类问题。:-)|1633779|entityMap|0|LINK|mutability|MUTABLE|url|http://www.ideone.com/smh73|1|http://loki-lib.sourceforge.net/^0|0|0|0|7|R|0|0|0|0|8|1|A|2|0|0|0|1M|K|0|0|4|1|0^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|1B|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|1C|8|@]|9|@]|A|$B|C]]|$1|I|3|J|5|F|7|1D|8|@]|9|@$K|1E|L|1F|1|1G]]|A|$]]|$1|M|3|N|5|F|7|1H|8|@]|9|@]|A|$]]|$1|O|3|-4|5|F|7|1I|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|1J|8|@$K|1K|L|1L|S|T]|$K|1M|L|1N|S|T]]|9|@]|A|$]]|$1|U|3|-4|5|F|7|1O|8|@]|9|@]|A|$]]|$1|V|3|-4|5|F|7|1P|8|@]|9|@]|A|$]]|$1|W|3|X|5|F|7|1Q|8|@$K|1R|L|1S|S|T]]|9|@]|A|$]]|$1|Y|3|Z|5|F|7|1T|8|@]|9|@$K|1U|L|1V|1|1W]]|A|$]]|$1|10|3|-4|5|F|7|1X|8|@]|9|@]|A|$]]]|11|$12|$5|13|14|15|A|$16|17]]|18|$5|13|14|15|A|$16|19]]]]

<pre><code>template &lt;class T&gt;
static T ( *pfunction ) ( const Object &lt;T&gt; *, const Object &lt;T&gt; *);
</code></pre>

Template of function pointer is illegal in C++. Be it inside a class, or simply outside a class. You cannot write this (not even outside a class):

<pre><code>template &lt;class X&gt;
void (*PtrToFunction) (X);
</code></pre>

See this sample : <a href="http://www.ideone.com/smh73" rel="noreferrer">http://www.ideone.com/smh73</a>

The C++ Standard says in $14/1,

<blockquote>
 A template defines a family of classes
 or functions.
</blockquote>

Please note that it does NOT say "A template defines a family of classes, functions or function pointers". So what you're trying to do is, defining "a family of function pointers" using template, which isn't allowed. 

Generic Functors from <a href="http://loki-lib.sourceforge.net/" rel="noreferrer">Loki</a> library would be an elegant solution to the kind of problem you're having. :-)

blocks|key|2033954|text|您可以做的一件事是在cpp文件中有一个模板成员函数的副本，并指向该函数，即|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2033955|template+<%2Btypename+ElementType>
int+PQueueHeap<ElementType>::compareFunction(ElementType+First,ElementType+Second)
{+++
++++if+(First>Second)+return+1;+else+if+(First==Second)+return+0;+else+return+-1;
}

//+you+cannot+point+to+above+|code-block|syntax|javascript|2033956|但是，您可以指向|2033957|template+<%2Btypename+ElementType>

int+compareFunction(ElementType+First,ElementType+Second)
{

if+(First>Second)+return+1;+else+if+(First==Second)+return+0;+else+return+-1;
}+//+No+error+and+it+works!+|2033958|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

One thing you can do is have a copy of the template member function in the cpp file and point to that i.e. 

<pre><code>template &lt;+typename ElementType&gt;
int PQueueHeap&lt;ElementType&gt;::compareFunction(ElementType First,ElementType Second)
{ 
 if (First&gt;Second) return 1; else if (First==Second) return 0; else return -1;
}

// you cannot point to above 
</code></pre>

however you can point to 

<pre><code>template &lt;+typename ElementType&gt;

int compareFunction(ElementType First,ElementType Second)
{

if (First&gt;Second) return 1; else if (First==Second) return 0; else return -1;
} // No error and it works! 
</code></pre>

I am having a pointer to the common static method

<pre><code>class MyClass
{
 private:
 static double ( *pfunction ) ( const Object *, const Object *);
 ...
};
</code></pre>

pointing to the static method

<pre><code> class SomeClass
 {
 public:
 static double getA ( const Object *o1, const Object *o2);
 ...
 };
</code></pre>

Initialization:

<pre><code>double ( *MyClass::pfunction ) ( const Object *o1, const Object *o2 ) = &amp;SomeClass::getA;
</code></pre>

I would like to convert this pointer to the static template function pointer:

<pre><code>template &lt;class T&gt;
static T ( *pfunction ) ( const Object &lt;T&gt; *, const Object &lt;T&gt; *); //Compile error
</code></pre>

where:

<pre><code> class SomeClass
 {
 public:
 template &lt;class T&gt;
 static double getA ( const Object &lt;T&gt; *o1, const Object &lt;T&gt; *o2);
 ...
 };
</code></pre>

But there is the following compile error: 

<pre><code>error: template declaration of : T (* pfunction )(const Object &lt;T&gt; *o1, const Object &lt;T&gt; *o2)
</code></pre>

Thanks for your help...

C++, function pointer to the template function pointer

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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