如果声明指针是void*，它意味着任何型的地址都可以间接引用那个指针
那么，在*((int *)p)=3中是不是在把空指针改了以后，成为整形的，就一直是整形的 不能改变吗 ？？？？

void* is called the generic pointer --- every pointer is 4 bytes (32-bit os) so that you can cast between different pointer types.

*((int*)p) = 3;

only says that the content of the memory location to which p points is 3.

It does not say p is a void*, or a int*.

哦好象说的有道理，那让我在仔细看看把
谢谢了！