XOR一个非常大的文件 [英] XOR on a very big file

问题描述

我想XOR一个非常大的文件（〜50 GO）。

I would like to XOR a very big file (~50 Go).

更多precisely，我想用钥匙3847611839异或32字节的纯文本文件的（因为内存不足），每块这样做，并创建（块之后块）一个新的密码文件。

More precisely, I would like to do so by XORing each block of 32 bytes of a plaintext file (because of lack of memory) with the key 3847611839 and create (block after block) a new cipher file.

感谢您的帮助！

推荐答案

这听上去很不错，并没有听起来像一个家庭作业。

This sounded like fun, and doesn't sound like a homework assignment.

我没有一个previously异或加密的文件，试图用，但如果你转换一个前进和后退，也没有差异。

I don't have a previously xor-encrypted file to try with,but if you convert one back and forward, there's no diff.

这是我试过AT​​LEAST。请享用！ :)这XOR的每一个4个字节0xE555E5BF，我presume这就是你想要的。

That I tried atleast. Enjoy! :) This xor's every 4 bytes with 0xE555E5BF, I presume that's what you wanted.

下面是bloxor.c

Here's bloxor.c

// bloxor.c - by Peter Boström 2009, public domain, use as you see fit. :)

#include <stdio.h>

unsigned int xormask = 0xE555E5BF; //3847611839 in hex.

int main(int argc, char *argv[])
{
    printf("%x\n", xormask);
    if(argc < 3)
    {
    	printf("usage: bloxor 'file' 'outfile'\n");
    	return -1;
    }

    FILE *in = fopen(argv[1], "rb");
    if(in == NULL)
    {
    	printf("Cannot open: %s", argv[2]);
    	return -1;
    }

    FILE *out = fopen(argv[2], "wb");

    if(out == NULL)
    {
    	fclose(in);
    	printf("unable to open '%s' for writing.",argv[2]);
    	return -1;
    }
    char buffer[1024]; //presuming 1024 is a good block size, I dunno...

    int count;

    while(count = fread(buffer, 1, 1024, in))
    {
    	int i;
    	int end = count/4;
    	if(count % 4)
    		++end;

    	for(i = 0;i < end; ++i)
    	{
    		((unsigned int *)buffer)[i] ^= xormask;
    	}
    	if(fwrite(buffer, 1, count, out) != count)
    	{
    		fclose(in);
    		fclose(out);

    		printf("cannot write, disk full?\n");

    		return -1;
    	}
    }

    fclose(in);
    fclose(out);

    return 0;
}

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