在C中打印地址或指针以获取值 [英] Print the address or pointer for value in C

问题描述

我想做一些看起来很简单的事情.我得到了结果，但是问题是，我无法知道结果是否正确.

I want to do something that seems fairly simple. I get results but the problem is, I have no way to know if the results are correct.

我正在C语言中工作，我有两个指针.我想打印指针的内容.我不想取消引用指针以获取指向的值，我只想要指针已存储的地址.

I'm working in C and I have two pointers; I want to print the contents of the pointer. I don't want to dereference the pointer to get the value pointed at, I just want the address that the pointer has stored.

我写了以下代码，我需要知道的是，如果正确，如何纠正.

I wrote the following code and what I need to know is if it is right and if not, how can I correct it.

/* item one is a parameter and it comes in as: const void* item1   */
const Emp* emp1 = (const Emp*) item1; 

printf("\n comp1-> emp1 = %p; item1 = %p \n", emp1, item1 );

尽管我要发布此内容(以及正确的理由很重要)，但我最终还是需要为指向指针的指针执行此操作.那就是:

While I'm posting this (and the reason it is important that it is correct) is that I eventually need to do this for a pointer-to-a-pointer. That is:

const Emp** emp1 = (const Emp**) item1; 

推荐答案

您所拥有的是正确的.当然，您会看到emp1和item1具有相同的指针值.

What you have is correct. Of course, you'll see that emp1 and item1 have the same pointer value.

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