小端或大端 [英] Little endian or Big endian

问题描述

#include <stdio.h>

union Endian
{
    int i;
    char c[sizeof(int)];
};

int main(int argc, char *argv[]) 
{
    union Endian e;
    e.i = 1;
    printf("%d \n",&e.i);
    printf("%d,%d,\n",e.c[0],&(e.c[0]));
    printf("%d,%d",e.c[sizeof(int)-1],&(e.c[sizeof(int)-1]));


}

输出:

1567599464 
1,1567599464,
0,1567599467

LSB存储在较低的地址中，而MSB存储在较高的地址中.这不是大端吗?但是我的系统配置将其显示为小端架构.

LSB is stored in the lower address and MSB is stored in the higher address. Isn't this supposed to be big endian? But my system config shows it as a little endian architecture.

推荐答案

您的系统肯定是little-endian.如果是big-endian，则以下代码:

You system is definitely little-endian. Had it been big-endian, the following code:

printf("%d,%d,\n",e.c[0],&(e.c[0]));

将为第一个%d而不是1打印0.在little-endian中，1存储为

would print 0 for the first %d instead of 1. In little-endian 1 is stored as

00000001 00000000 00000000 00000000
^ LSB
^Lower Address

，但在big-endian中，它存储为

00000000 00000000 00000000 00000001
                           ^LSB
                           ^Higher Address  

并且不要使用%d打印变量的地址，请使用%p.

And don't use the %d to print addresses of variables, use %p.

这篇关于小端或大端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！