在C中取消引用

Question

我刚刚开始学习C所以请善待.从我到目前为止读到的关于指针的内容:

int * test1; //this is a pointer which is basically an address to the process 
             //memory and usually has the size of 2 bytes (not necessarily, I know)
float test2; //this is an actual value and usually has the size of 4 bytes,
             //being of float type
test2 = 3.0; //this assigns 3 to `test2`

现在,我不完全理解:

*test1 = 3; //does this assign 3 at the address 
            //specified by `pointerValue`?
test1 = 3;  //this says that the pointer is basically pointing 
            //at the 3rd byte in process memory, 
            //which is somehow useless, since anything could be there
&test1; //this I really don't get, 
        //is it the pointer to the pointer? 
        //Meaning, the address at which the pointer address is kept?
        //Is it of any use?

同理:

*test2; //does this has any sense?
&test2; //is this the address at which the 'test2' value is found? 
        //If so, it's a pointer, which means that you can have pointers pointing 
        //both to the heap address space and stack address space. 
        //I ask because I've always been confused by people who speak about 
        //pointers only in the heap context.

Answer 1

很好的问题。

你的第一个块是正确的。指针是保存某些数据地址的变量。该指针的类型告诉代码如何解释该指针所保存的地址的内容。

构造：

*test1 = 3

称为指针的引用。这意味着，您可以像普通变量一样访问指针指向的地址并对其进行读写。笔记：

int *test;
/*
 *    test is a pointer to an int - (int *)
 *   *test behaves like an int - (int)
 *
 * So you can thing of (*test) as a pesudo-variable which has the type 'int' 
 */

以上只是我使用的一个助记装置。

您很少为指针分配数值...也许如果您正在为具有一些“众所周知”内存地址的特定环境进行开发，但在您的级别，我不会太担心那。

使用

*test2

最终会导致错误。您会尝试遵循不是指针的东西，因此您可能会遇到某种系统错误，因为谁知道它指向哪里。

&test1and&test2实际上是指向test1和 的指针test2。

指向指针的指针非常有用，搜索指向指针的指针将引导您找到一些比我更好的资源。