取数组地址时不同的指针算术结果 [英] Different Pointer Arithmetic Results when Taking Address of Array
本文介绍了取数组地址时不同的指针算术结果的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
程序:
#include<stdio.h>
int main(void) {
int x[4];
printf("%p\n", x);
printf("%p\n", x + 1);
printf("%p\n", &x);
printf("%p\n", &x + 1);
}
输出:
$ ./a.out
0xbff93510
0xbff93514
0xbff93510
0xbff93520
$
我希望以下是上述程序的输出.例如:
I expect that the following is the output of the above program. For example:
x // 0x100
x+1 // 0x104 Because x is an integer array
&x // 0x100 Address of array
&x+1 // 0x104
但是最后一条语句的输出与我期望的不同. &x
也是数组的地址.所以在这个上增加1
将打印以4递增的地址.但是&x+1
给出以10递增的地址.为什么?
But the output of the last statement is different from whast I expected. &x
is also the address of the array. So incrementing 1 on this
will print the address incremented by 4. But &x+1
gives the address incremented by 10. Why?
推荐答案
x -> Points to the first element of the array.
&x ->Points to the entire array.
Stumbled upon a descriptive explanation here: http://arjunsreedharan.org/post/69303442896/the-difference-between-arr-and-arr-how-to-find
SO链接:为什么arr和& arr相同?
这篇关于取数组地址时不同的指针算术结果的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
查看全文