用C交换字节序 [英] Swapping endiannes in C

问题描述

我有这个字符串

c1eb044f0708015b267913fc4dff5aabe3dd4a97f10f7ba935cd360000000000

一个人怎么掉它，所以它成为

How does one swap it so it becomes

000000000036cd35a97b0ff1974adde3ab5aff4dfc1379265b0108074f04ebc1

这两个基本的例子，但是这就是我需要做的，但不知道怎么，我对C知之甚少。

Those two are basically examples, but that is what i need to do, but not know how as i have very little knowledge of C.

以上两个字符串实际上是无符号的字符[]在C程序

The above two strings are actually unsigned char[] in the C program

P.S
别以为我没有去通过谷歌。我这样做，但我发现很少的东西，我需要让每一个试图这样做失败了。

P.S Don't think i didn't go through google. I did, but i found very little of what i needed so every attempt to do that failed.

推荐答案

像这样的事情;也许并不完美，但给你的想法。你要适当的错误检查，您的缓冲区的初始化等。

Something like this; probably not perfect but gives you the idea. You'll want appropriate error checking, initialization of your buffer, etc.

编辑：我做了一个假设，我不应该，有可能。我间preTED你的字符串作为十六进制重新字节presentations，所以我把 C1 为 unsigned char型和 00 ，例如。如果字符串实际上是小写字母c，1号等，然后弥的回答是你想要的，不是我的。

I made an assumption I shouldn't have, possibly. I interpreted your string as hex representations of bytes, so I took c1 as an unsigned char and switched it with 00, for example. If your string is actually lowercase c, the number 1, etc., then Micah's answer is what you want, not mine.

void reverse_string(unsigned char *buf, int length)
{
    int i;
    unsigned char temp;

    for (i = 0; i < length / 2; i++)
    {
        temp = buf[i];
        buf[i] = buf[length - i - 1];
        buf[length - i - 1] = temp;
    }   
}

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