My function not working 'name.exe has stopped working'

Question

The program stops and I get a "merge.exe has stopped working" message. I simplefied everything but the problem seems to occur when calling the function...

#include <stdio.h>
#include <stdlib.h>
#define br 1000
int sort(int *list[],int broi);
int merge(int list1[],int list2[],int broi1,int broi2);
int main()
{
int i,n,m,a1[br],a2[br],a;
printf("gimme n: ");
scanf("%d",&n);
printf("gimme m: ");
scanf("%d",&m);
for(i=0;i<n;i++)
{
    printf("gimme element %d ot list1: ",i+1);
    scanf("%d",&a1[i]);
}
for(i=0;i<m;i++)
{
    printf("gimme element %d ot list2: ",i+1);
    scanf("%d",&a2[i]);
}
sort(&a1,n);
sort(&a2,m);
merge(a1,a2,n,m);
return 0;
}
int sort(int *list[],int broi)
{
int i,j,c;
for (i = 0; i < broi-1; i++)
{
    for (j = 0; j < broi-i-1; j++)
    {
        if (*list[j] > *list[j+1])
        {
            c=*list[j];
            *list[j]=*list[j+1];
            *list[j+1]=c;
        }
    }
}
return 1;
}
int merge(int list1[],int list2[],int broi1,int broi2)
{
printf(" AAAAAA");
return 1;
}

In the sort() function is pretty self-explanatory and it even works! But whatever I put in merge() it just won't won't work and it breaks the whole program.

Answer 1

In the sort() function is pretty self-explanatory and it even works! ? NO it won't work. Read the compiler warning by compiling with -Wall it says

reo.c:4:5: note: expected ‘int **’ but argument is of type ‘int *’ as you are passing address of a1 & catching with array of pointer. Instead just pass array name and catch with pointer.

Make function call looks like

sort(a1,n);/*just pass the array name */
sort(a2,m);

And definition of sort()

int sort(int *list,int broi){ /* take list is ptr variable */
        int i,j,c;
        for (i = 0; i < broi-1; i++) {
                for (j = 0; j < broi-i-1; j++) {
                        if (list[j] > list[j+1]) {
                                c=list[j];
                                list[j]=list[j+1];
                                list[j+1]=c;
                        }
                }
        }
}

Answer 2

int *list[] makes list an array of pointers to int. In your main function &a1 (for example) create a pointer to an array of int, of type int (*)[1000]. int** (from the first) is not the same as int (*)[1000].

Simple solution? Don't pass a pointer to the array! This is very seldom needed.

And anyway because of operator precedence an expression like *list[i] is the same as *(list[i]), and if you have a pointer to an array you want (*list)[i]. So, many wrongs don't make a right.

Your compiler should have been shouting warnings about this to you.