Illegal initialization while trying to compare two groups of integers using pointers

[WraithBug]

Hello everyone. Well, being a total n00b, I know I pretty much suck at this for now while I'm learning. I'm trying to compare two groups of integers and count the number of matches between the groups. With the current code I'm getting an "illegal initialization" error on line 8. I'm not yet able to see exactly where I went wrong though. Can you guys help me out with it please.

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


int main() 
{
    int sysHunds,sysTens,sysOnes,usrHunds,usrTens,usrOnes;
    int * drawSet[3]={&sysHunds,&sysTens,&sysOnes};
    int * usrSet[3]={&usrHunds,&usrTens,&usrOnes}; 
    int ds, us, matches=0;
    
        for (ds=0;ds<3;ds++) for (us=0;us<3;us++) 
        if (*drawSet[ds]==*usrSet[us]) matches++;
        printf("\nThere are %d matching number(s)!\n", matches);
        
    
system("pause");
return 0;
}

[laserlight]

I think you mean line 9 rather than line 8. Basically, prior to C99 (according to my compiler), you are not permitted to initialise an array like that, i.e., by taking the addresses of variables. If feasible, a simple solution is just to compile as C99, otherwise you would change to write:

int *drawSet[3];
drawSet[0] = &sysHunds;
/* etc */

I recommend having two arrays of ints instead:

int drawSet[3];
int usrSet[3];

If you want to refer to different elements by name, use an enum for the indices.

[WraithBug]

Thanks for the help @laserlight. I appreciate it and I went with your first option (although I may change it to the arrays later) and it's almost perfect. The only problem now is that when either *drawSet or *usrSet has a number where the same digit appears twice (ex: 454) and it appears once in the other set (ex: 348) it counts it as two matches. The goal is to count once for each cross matched pair, not counting the same occurrence of a digit for a second time. Can this be done with changes to the current code or might there be an easier way to accomplish it?

Here's the code again with printf of the above stated results:

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


int main() 
{
    int sysHunds=4,sysTens=5,sysOnes=4,usrHunds=3,usrTens=4,usrOnes=8,ds,us,matches=0;
    int * drawSet[3]; int * usrSet[3]; 
    
    drawSet[0] = &sysHunds;
    drawSet[1] = &sysTens;
    drawSet[2] = &sysOnes;
    usrSet[0] = &usrHunds;
    usrSet[1] = &usrTens;
    usrSet[2] = &usrOnes;
    
    printf("\n*drawSet: %d%d%d        usrSet: %d%d%d\n", *drawSet[0], *drawSet[1], *drawSet[2], *usrSet[0], *usrSet[1], *usrSet[2]);
    
        for (ds=0;ds<3;ds++) for (us=0;us<3;us++) 
        if (*drawSet[ds]==*usrSet[us]) matches++;
        
        printf("\nThere are %d matching number(s)!\n", matches);


    
system("pause");
return 0;
}

[laserlight]

The only problem now is that when either *drawSet or *usrSet has a number where the same digit appears twice (ex: 454) and it appears once in the other set (ex: 348) it counts it as two matches. The goal is to count once for each cross matched pair, not counting the same occurrence of a digit for a second time. Can this be done with changes to the current code or might there be an easier way to accomplish it?

One approach is to sort both arrays and then step through them in a single pass each.

[WraithBug]

One approach is to sort both arrays and then step through them in a single pass each.

Thanks again. I'll try to get that done and hopefully see what happens.