模板中的C ++函数指针 [英] C++ function pointers inside templates

问题描述

我有这个问题:

template<typename T> class Bubu
{
...
   int (*comparer)(const T t1, const T t2);
...
public:
   Bubu(int (*_comparer)(const T t1, const T t2))
   {
      comparer = _comparer;
   }
};

在另一个文件中:

Bubu<char*> asd(strcmp);

错误:

error C2664: 'Bubu<T>::Bubu(int (__cdecl *)(const T,const T))' : 
             cannot convert parameter 1 from 'int (__cdecl *)(const char *,
             const char *)' to 'int (__cdecl *)(const T,const T)'

我不明白为什么.

支持Ideone.com的代码:

the Ideone.com-ready code:

int asdf(const char* a, const char* b)
{       return 0; }

template class Bubu
{
   int (*comparer)(const T t1, const T t2);
public:
   Bubu(int (*_comparer)(const T t1, const T t2))
   {
      comparer = _comparer;
   }
};

int main(int argc, char* argv[])
{
Bubu asd(asdf);
}

推荐答案

当T为char*时，const T为char* const，与const char *不同.您需要:

When T is char*, const T is char* const which isn't the same thing as const char *. You need:

 Bubu<const char*> asd(strcmp);

对于函数签名，顶层const被忽略

Top level const is ignored for function signatures so

int (*)( const char* const, const char* const );

与

int (*)( const char*, const char* );

尽管在简单的int (*comparer)(T t1, T t2);上并不能给您带来任何好处，但是您可以在顶级顶级常量上使用.

so you're OK on the extra top level const although it doesn't gain you anything over the simpler int (*comparer)(T t1, T t2);.

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