什么是C ++中的void指针？ [英] What is a void pointer in C++?

问题描述

可能重复：

我经常看到类似下面的代码：

I often see code which resembles something like the following:

void * foo int bar）;

这是什么意思？这是否意味着它可以返回任何 ？这类似于C＃中的动态或对象吗？

What does this mean? Does it mean that it can return anything? Is this similar to dynamic or object in C#?

推荐答案

A void * 并不意味着什么。它是一个指针，但它指向的类型是未知的。

A void* does not mean anything. It is a pointer, but the type that it points to is not known.

这不是它可以返回任何东西。返回 void * 的函数通常执行以下操作之一：

It's not that it can return "anything". A function that returns a void* generally is doing one of the following:

• 它正在处理未格式化的内存。这是 operator new 和 malloc 返回：指向一定大小的内存块的指针。因为内存没有类型（因为它还没有一个正确构造的对象），它是无类型的。 IE： void 。


• 这是一个不透明的句柄;它引用创建的对象而不命名特定类型。这样做的代码通常形式不好，因为这最好通过向前声明一个结构/类，而不是为它提供一个公共定义。


• 它有明确的文档告诉你可以使用指针的类型。

• It is dealing in unformatted memory. This is what operator new and malloc return: a pointer to a block of memory of a certain size. Since the memory does not have a type (because it does not have a properly constructed object in it yet), it is typeless. IE: void.
• It is an opaque handle; it references a created object without naming a specific type. Code that does this is generally poorly formed, since this is better done by forward declaring a struct/class and simply not providing a public definition for it. Because then, at least it has a real type.
• It has explicit documentation telling you what type(s) that you can use the pointer for.

不是动态或物件在C＃。这些结构实际上知道什么是原始类型; void * 不会。这使得它比任何那些更危险，因为它很容易弄错。

It is nothing like dynamic or object in C#. Those constructs actually know what the original type is; void* does not. This makes it far more dangerous than any of those, because it is very easy to get it wrong.

在个人注意，如果你看到使用 void * 的often，你应该重新思考你正在看什么代码。 void * 用法，特别是C ++中的，应该是罕见的，用于处理原始内存。

And on a personal note, if you see code that uses void*'s "often", you should rethink what code you're looking at. void* usage, especially in C++, should be rare, used primary for dealing in raw memory.

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