C：指针运算

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我是新的C语言编程，并试图了解指针运算工作。下面的printf语句打印2时printf的参数是*（P + 2）和4与* P。能否请您解释这种现象？

的#include＆LT;＆stdio.h中GT;＃包括LT＆;＆CONIO.H GT;诠释的main（）{    INT ARR [4] = {4,3,2,1}，* P =改编;    的printf（\\ N％D，*（P + 2））;    返回0;}

解决方案

让我们重新编写程序，以使其更清晰一点：

＃包括LT＆;＆stdio.h中GT;INT主要（无效）{    INT ARR [4] = {4,3,2,1};    为int * p =改编;    的printf（\\ N％D，*（P + 2））;    返回0;}

现在， *（P + 2）是定义相同的 P [2] 。由于 P 点的ARR的第一个元素，那么 P [2] 相同改编[2] 等于 2 。

同样， *（P）相同 * P ，自点改编点的第一个元素，则 *（p）是 4 。

您可能需要重新读你的文字书，涵盖指针运算的部分。

I'm new to C programming and trying to understand how pointer arithmetic works. The below printf statement prints 2 when the arguments for printf is *(p+2) and 4 with for *p. Could you please explain this behaviour ?

#include <stdio.h> #include <conio.h> int main() { int arr[4] = {4,3,2,1}, *p = arr; printf("\n%d", *(p+2)); return 0; }

解决方案

Let's re-write your program to make it a little clearer:

#include<stdio.h> int main(void) { int arr[4] = {4,3,2,1}; int *p = arr; printf("\n%d", *(p+2)); return 0; }

Now, *(p+2) is by definition the same as p[2]. Since p points to the first element of arr, then p[2] is the same as arr[2] which is equal to 2.

Similarly, *(p) is the same as *p and since p points to the first element of arr then *(p) is 4.

You probably need to re-read the section in your text book that covers pointer arithmetic.