C和C ++中的空指针之间的差异 [英] Difference between void pointers in C and C++

问题描述

为什么C ++中的下列错误（但在C中有效）

Why the following is wrong in C++ (But valid in C)

void*p;
char*s;
p=s;
s=p; //this is wrong ,should do s=(char*)p;

为什么需要转换， p 现在包含 char 指针和 s 的地址也是 char 指针？

Why do I need the casting,as p now contains address of char pointer and s is also char pointer?

推荐答案

值不重要，类型是。由于 p 是一个void指针， s 一个字符指针，你必须转换，即使它们具有相同的值。在C中它会确定， void * 是通用指针，但在C ++中是不正确的。

The value doesn't matter, the type does. Since p is a void pointer and s a char pointer, you have to cast, even if they have the same value. In C it will be ok, void* is the generic pointer, but this is incorrect in C++.

p 不包含char指针，它是一个void指针，它包含一个内存地址。

By the way, p doesn't contains char pointer, it's a void pointer and it contains a memory address.

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