CODEWITHSUNDEEP
cencryptionxor
vikkyhacks
vikkyhacks

Reputation: 3230

Extra 0xFFFFFF during xor encryption

I have this piece of code here

#include<stdio.h>

#define LEN 10

char buf[] = {0xff,0xaa,0xfc,0xe8,0x89,0x00,0x00,0x00,0x60,0x89};
char key[] = "SAMPLE KEY";

int main()
{
        int i;
        char enc[LEN];
        for(i=0;i<LEN;i++)
        {
                enc[i] = (key[i % LEN] ^ buf[i]);
                printf("0x%x,",enc[i]);
        }
}

Which outputs

OUTPUT

0xffffffac,0xffffffeb,0xffffffb1,0xffffffb8,0xffffffc5,0x45,0x20,0x4b,0x25,0xffffffd0

where the expected output is

0xac,0xeb,0xb1,0xb8,0xc5,0x45,0x20,0x4b,0x25,0xd0,

How can I fix this ? Why does this happen ?

Upvotes: 1

Views: 387

Answers (4)

chux
chux

Reputation: 154085

OP is using a format meant %x for unsigned.

Since a char value becomes an int with printf() ... parameters, use a format %hhx meant for char. This will print only the char portion. Example:

printf("0x%02hhx,",enc[i]);  --> 0xac,

Upvotes: 0

Ilmari Karonen
Ilmari Karonen

Reputation: 50368

Your encryption works fine. You problem is that you've declared enc as an array of signed chars. (As ouah correctly notes, char can be either signed or unsigned by default. On your compiler, it's apparently signed.)

Thus, chars like 0xAC which have their MSB set are interpreted as negative numbers. When you pass them to printf(), they're implicitly converted to ints, and therefore sign-extended to 32 bits.

To fix it, just declare enc, key and buf explicitly as arrays of unsigned chars. Alternatively, you could use Ben Voigt's solution and mask off the extra bits, but then you'd be pointlessly sign-extending your chars just to mask those extra bits off again.

Ps. You do realize that this kind of "encryption" is hopelessly weak if you're encrypting anything significantly longer than the key, right?

Upvotes: 1

Ben Voigt
Ben Voigt

Reputation: 283773

Right now, you not only have leading sign-extended bits, but also missing leading zeros.

Use

printf("0x%02x,", enc[i] & 0xFFu);

Upvotes: 0

ouah
ouah

Reputation: 145899

Change buf and enc type at declaration from char to unsigned char.

It is implementation-defined whether char is a signed or unsigned type and in your implementation it is apparently a signed type.

Upvotes: 3

Related Questions