C language-pointer arithmetic example

/*指针运算示例*/
#include<stdio.h>
int main()
{
    
    
	int i, a[] = {
    
     1,2,3,4,5 }, * p = a;
	for (i = 0; i < 5; i++)
		printf("%d\n", a[i]);          //循环输出a[i]的值
	printf("a = %d\n", p);             //输出数组a的首地址,即a[0]的地址
	printf("p + 2 = %d\n", p + 2);     //输出a[0]的首地址,然后+2。因为整型占用四个字节,所以结果应为a[0]的首地址值+8
	printf("* p + 3 = %d\n", * p + 3); //输出a[0]+3,因为p指向a[0]的首地址,所以*p即a[0],*p+3的值等于a[0]+3
	printf("* (p + 3) = %d\n", * (p + 3));   //输出a[0+3]=a[3]的值
	printf("* p++ = %d\n", * p++);     //输出a[0]的值,然后指针指向a[1]。“*p++” 等价于“ *(p++)”
	p = a;                             //令p重新指向a[0]
	printf("* ++p = %d\n", * ++p);     //“* ++p”等价于“* (++p)”,指针先从指向a[0]变到指向a[1],然后输出a[1]的值,注意区分和* p++的区别
	printf("++ *p = %d\n",++ *p);      // “++ *p”等价于 “++(*p)”,++作用于“*p”,因为此时p指向a[1],所以先进行a[1]+1,然后再输出a[1]+1
	return 0;
}

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Origin blog.csdn.net/qq_43516928/article/details/114780902