C ++有可能有一个通用的函数指针吗？ [英] C++ Is it possible to have a generic function pointer?

问题描述

在C ++中，有可能做一些类型的通用函数指针，指向返回一个类型的指针并且不带参数的任何函数。

In C++ is it possible to make some sort of generic function pointer that points to any function that returns a pointer to some type and takes no arguments?

一种可以指向以下两种类型的指针：

Eg, one type of pointer that can point to both of the following:

int* funcInt(){
    int* i = new int;
    *i = 5;
    return i;
}

char* funcChar(){
    char* c = new char;
    *c = 'a';
    return c;
}

显然以下是有效的：

int* (*funcPointerA)() = funcInt;
char* (*funcPointerB)() = funcChar;

但是可以做类似下面的事情（它给出了编译错误）

But is it possible to do something like the following (it gives a compile error at the moment):

void* (*funcPointerC)() = funcInt;
void* (*funcPointerD)() = funcChar;

目前上述代码给出以下错误：

Currently the above code gives the following error:

error: invalid conversion from 'int* (*)()' to 'void* (*)()'
error: invalid conversion from 'char* (*)()' to 'void* (*)()'

funcPointerA和B到C和D？

Is there any way to cast the funcPointerA and B to C and D?

这样就可以将指针指向两种类型的函数（以及在没有参数的情况下返回指向某种类型的指针的指针）可以一起存储在单个向量中。 p>

This is so pointers to both types of functions (and others that return a pointer to some type while taking no arguments) can be stored together in a single vector.

推荐答案

标准不允许这样的事情。你可能也可能不能把函数指针转换为 void *（*）（）。在C ++中有符合标准的解决方案。这里是一个简单的前C ++ 11一个：

The standard does not allow such thing. You may or may not be able to get away with casting your function pointer to void* (*)(). In C++ there are standard-conforming solutions. Here is a simple pre-C++11 one:

struct MyFunc
{
  virtual void* operator()() = 0;
  virtual ~Myfunc(){}
};

template <typename T>
struct MyfuncImpl
{
  typedef T TFunc();
  MyfuncImpl (TFunc* func) : m_func(func) {}
  void* operator()() { return m_func(); }
 private:
  TFunc* m_func;
};

现在您可以存储 shared_ptr< Myfunc>

在C ++ 11中更好的解决方案可能如下所示：

A much nicer solution in C++11 could look like this:

template <typename T>
std::function<void*()> castToVoidFunc (T* (*func)())
{
  return [=](){ return func(); };
}

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