Solution for Exercise 20-7

ex2007

#include <stdio.h>
#include <stdlib.h>

int main()
{
    unsigned char *junk;
    int x;

    /* allocate memory */
    junk = (unsigned char *)calloc(64,sizeof(char));
    if( junk==NULL )
    {
        puts("Unable to allocate memory");
        exit(1);
    }

    /* output the buffer */
    for(x=0;x<64;x++)
    {
        printf("%02X ",*(junk+x));
        if( (x+1) % 8 == 0 )
            putchar('\n');
    }

    return(0);
}

Output

00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00
00 00 00 00 00 00 00 00

Notes

* Unlike Exercise 20-6, the output for this code is always all zeros.

* I added a typecast to the calloc() function. As the junk pointer is an unsigned char, calloc() is typecast from void to an unsigned char pointer.

Exercise ex2008 →