Solution for Exercise 17-9

ex1709

#include <stdio.h>

char *binbin(unsigned char n);

int main()
{
    int a,x,r;

    a = 73;
    x = 170;

    printf("  %s %3d\n",
            binbin((unsigned char)a),a);
    printf("^ %s %3d\n",
            binbin((unsigned char)x),x);
    r = a ^ x;
    printf("= %s %3d\n",
            binbin((unsigned char)r),r);
    printf("^ %s %3d\n",
            binbin((unsigned char)x),x);
    a = r ^ x;
    printf("= %s %3d\n",
            binbin((unsigned char)a),a);
    return(0);
}

char *binbin(unsigned char n)
{
    static char bin[9];
    int x;

    for(x=0;x<8;x++)
    {
        bin[x] = n & 0x80 ? '1' : '0';
        n <<= 1;
    }
    bin[x] = '\0';
    return(bin);
}

Output

01001001 73
^ 10101010 170
= 11100011 227
^ 10101010 170
= 01001001 73

Notes

* In Line 21 I'm re-using the a variable to avoid having to declare another variable simply to hold the result of the XOR operator. Even so, due to XOR's charm, the value that ends up in variable a is the same value it began with, 73.

* The XOR operator can be used as a quick and dirty cipher. The code is easy to break, so it's by no means secure, but the ^ operator can be used to swiftly encrypt and decrypt text. You can try coding such a program on your own; click here for my example.

Exercise ex1710 →