做两个结构相同的指针 [英] making two struct pointers equal

问题描述

所以我们可以说我有一个结构的指针称为A和结构指针称为B：

So let's say i have a struct pointer called A and struct pointer called B:

struct example{
//variables and pointers
}*A 

然后我有数据类型结构示例的指针：

and then i have a pointer of datatype struct example:

struct example *B=malloc(sizeof(struct example));

如果我这样做

A=B;

这是否运算意味着什么结构指针B的指向是什么结构指针A也指向？我得到它与原始数据类型和指针，但结构迷惑我，因为他们里面..

does this arithmetic operation mean that whatever struct pointer B is pointing to will be what struct pointer A is also pointing to? I get it with primitive datatype and pointers but struct confuse me because they have variables inside..

假设结构指针设置，一切

assume struct pointer A is set and everything

推荐答案

是与code以下线，应指向同一位置，B被指向。

Yes with the following lines of code, A should point to the same location, B was pointing to.

/* Define the struct example datatype */
struct example{
    //variables and pointers
};

/* Declare A and B as pointers to struct example */
struct example *A;
struct example *B;

/* Allocate memory equivalent to the size of struct example
   and store the address in B */
B=malloc(sizeof(struct example));

/* Copy the address in B to A, so that A points to the same
   location B was pointing to */
A = B;

您应该把指针只是另一种无符号长变量，它拥有一个地址到内存，（因为指针正好指向其持有的地址）。

You should think of pointers as just another unsigned long variable which holds an address to a memory, (since the pointer just points to the address it holds).

正如任何其他变量，可以存储在一个指针变量的地址复制到另一个。

Just as with any other variables, you can copy the address stored in one pointer variable to another.

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