Question

The code is just very simple, yet I scratch my head at the results.
I am just playing pointer arithmetics and want to print out the array
but I get the numbers of the array plus 3 more. Where do those 3 extra come from ?

#include <stdio.h>


int my_array[] = {1,3,5,6,73,343,34};

int *pointer_numeros;

int main (void) {
int i = 0;  
pointer_numeros = my_array;

while(*pointer_numeros) {

        printf("los numeros del array son %d\t %d\n\n", i++, *pointer_numeros++);

            }

            getchar();
            return 0;

    }   

Answer 1

*pointer_numeros does not evaluate to false at the end of the array; it will carry on walking through memory until it hits an address whose contents are zero (but this is undefined behaviour).

You can terminate your array in a zero, as others have suggested. But in general, you will still have a problem: what if some of your elements are themselves zero?