Question

I have this piece of code in C:

int x = 52706108;

 if(argc >= 2){
  int val = *argv[1];
  int xor = x^val;
  printf("The xor value between %d and %d is %d in decimal\n",x,val,xor);
 }

I’m compiling it like this:

gcc -m32 -g -o a5_1 a5_1.c

Running it like this:

./a5_1 12

And this is my output:

The xor value between 52706108 and 49 is 52706061 in decimal

I can’t understand why I’m passing the parameter “12” but the machine is reading 49 instead.

Answer 1

That 49 is the ASCII code point of the 1 in your string argument 12. That’s because argv is an array of char pointers, each of which points to a C string containing the argument. So, it’s as if you’ve defined argv[1] to be {'1', '2', '\0').

If you want to convert the argument to an integer, use something like:

int num = atoi (argv[1]);

or, preferably with error checking and to avoid undefined behaviour in the event the number is out of range:

char *nextChar;
long num = strtol (argv[1], &nextChar, 10);
if ((nextChar == argv[1]) || (*nextChar != '\0')) {
    // Is either empty or has invalid characters.
    return -1;
}

// String was non-empty and all-numeric.

---

Full example:

#include <stdio.h>
#include <stdlib.h>
int main (int argc, char *argv[]) {
    long x = 52706108;
    if (argc >= 2) {
        char *nextChar;
        long val = strtol (argv[1], &nextChar, 10);
        if ((nextChar == argv[1]) || (*nextChar != '\0')) {
            printf ("Invalid input '%s'\n", argv[1]);
            return -1;
        }
        long xor = x^val;
        printf("Xor between %ld and %ld is %ld in decimal\n",x,val,xor);
    }
    return 0;
}

The output of that program (when given 12 as an argument) is:

Xor between 52706108 and 12 is 52706096 in decimal