如何检查一个void *指针是否可以安全地转换成别的东西？ [英] How to check if a void* pointer can be safely cast to something else?

问题描述

假设我有这个函数，它是一些gui工具包的一部分：

Let's say I have this function, which is part of some gui toolkit:

typedef struct _My_Struct My_Struct;
/* struct ... */

void paint_handler( void* data )
{
   if ( IS_MY_STRUCT(data) ) /* <-- can I do something like this? */
   {
      My_Struct* str = (My_Struct*) data;
   }
}

/* in main() */
My_Struct s;
signal_connect( SIGNAL_PAINT, &paint_handler, (void*) &s ); /* sent s as a void* */

由于paint_handler也会被GUI工具包主循环与其他参数，我不能总是确保我接收的参数将永远是一个指针 s 。

Since the paint_handler will also be called by the GUI toolkit's main loop with other arguments, I cannot always be sure that the parameter I am receiving will always be a pointer to s.

我可以在 paint_handler 函数中执行 IS_MY_STRUCT ，以检查我接收的参数是否可以安全回到 My_Struct * ？

Can I do something like IS_MY_STRUCT in the paint_handler function to check that the parameter I am receiving can be safely cast back to My_Struct* ?

推荐答案

> void 指针会丢失其所有类型信息，因此，单独，您不能检查是否可以安全地转换。由程序员知道是否可以将 void * 安全地转换为类型。

Your void pointer looses all its type information, so by that alone, you cannot check if it can be cast safely. It's up to the programmer to know if a void* can be cast safely to a type.

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