Thread: passing 2d arrays as parameters

not able to understand where the problem is ..
Determining if a mtrix has in it's diagonal - all zeros

Code:

#include<stdio.h>

int isdiagzero(int a[][10],int m,int n,int *i,int *j)
{
    if(((*i)<m)&&((*j)<n))
    {
    if(a[*i][*j]!=0)
    	return 0;
   	else
   	{
   	    isdiagzero(a,m,n,i++,j++);
    }
    }    
    return 1;
}

int readmatrix(int a[][10],int m ,int n)
{
	int i,j;
    printf("Enter the row and column ");
    scanf("%d%d",&m,&n);
    if((m==0)||(n==0))
    	return 0;
   	else
   	{
   	    printf("\nEnter the matrix\n");
   		for(i=0;i<m;i++)
  			for(j=0;j<n;j++)
  				scanf("%d",&a[i][j]);
  	}
}

void printmatrix(int a[][10],int m ,int n)
{
    int i,j;
    printf("the matrix\n");
	for(i=0;i<m;i++)
	{
		for(j=0;j<n;j++)
			printf("%d",a[i][j]);
		printf("\n");
	}		
}

int main()
{
	int array[10][10];
	int m,n;
	
	if(readmatrix(array,m,n)==0)
		printf("ERROR\n");
	else
	{
	    printmatrix(array,m,n);
	    if(isdiagzero(array,m,n,0,0)==0)
	    	printf("Diagonal not zero !\n");
    	else
    		printf("Diagonal is zero !\n");
	}
}

this program reads input as
Enter the row and column 3 3
0 1 1
1 0 1
1 1 0
the matrix
----here it gives exception----
i am not getting it --
please help me...

> if(readmatrix(array,m,n)==0)
Should be
if(readmatrix(array,&m,&n)==0)

> int readmatrix(int a[][10],int m ,int n)
int readmatrix(int a[][10],int *m ,int *n)

> scanf("%d%d",&m,&n);
scanf("%d%d",m,n);

> for(i=0;i<m;i++)
for(i=0;i<*m;i++)
ditto for the other loop as well.

You need to return the size of the matrix as well, so you need to pass pointers to where the size will be stored.

i tried that but the exception is still coming
the output is correct if i use type-iteration
and an exception is generated from the type-recursive

the modified code is

Code:

#include<stdio.h>
int is = 0;
int isdiagzero(int a[][10],int m,int n,int *i,int *j)
{
    if(((*i)<m)&&((*j)<n))
    {
    if(a[*i][*j]!=0)
    	return 0;
   	else
   	{
   	    is = isdiagzero(a,m,n,i++,j++);		//is this parameter passing right i++,j++
    }
    }    
    return is;
}

int readmatrix(int a[][10],int *m ,int *n)
{
	int i,j;
    printf("Enter the row and column ");
    scanf("%d%d",m,n);
    if((*m==0)||(*n==0))
    	return 0;
   	else
   	{
   	    printf("\nEnter the matrix\n");
   		for(i=0;i<*m;i++)
  			for(j=0;j<*n;j++)
  				scanf("%d",&a[i][j]);
  	}
  	return 1;
}

void printmatrix(int a[][10],int m ,int n)
{
    int i,j;
    printf("the matrix\n");
	for(i=0;i<m;i++)
	{
		for(j=0;j<n;j++)
			printf("%d ",a[i][j]);
		printf("\n");
	}		
}

int isdiagz(int a[][10],int m ,int n)
{
    int i,j;
    printf("inside isdiag\n");
	for(i=0;i<m;i++)
		for(j=0;j<n;j++)
			if((i==j)&&(a[i][j]!=0))
				return 0;
	return 1;
}

int main()
{
	int array[10][10];
	int m,n;
	
	if(readmatrix(array,&m,&n)==0)
		printf("ERROR\n");
	else
	{
	    printmatrix(array,m,n);
	    if(isdiagzero(array,m,n,0,0)==0)	//type - recursion
//		if(isdiagz(array,m,n)==0)			//type - iteration
	    	printf("Diagonal not zero !\n");
    	else
    		printf("Diagonal is zero !\n");
	}
	getch();
	return 0;
}

this program reads input as
Enter the row and column 3 3
0 1 1
1 0 1
1 1 0
the matrix
0 1 1
1 0 1
1 1 0
----here it gives exception----
i am not getting it --
i suppose the problem lies in understanding recursion..
please help me...

> if(isdiagzero(array,m,n,0,0)==0)
But the last 2 parameters are pointers, so these are NULL pointers
No joy when you dereference them

Code:

int isdiagzero(int a[][10], int m, int n, int i, int j)
{
    if ((i < m) && (j < n)) {
        if (a[i][j] != 0)
            return 0;
        else {
            return isdiagzero(a, m, n, i+1, j+1);
        }
    }
    return 0;
}

i had understood that before
and tried writing the code as

Code:

	    int i=0,j=0;
	    if(isdiagzero(array,m,n,&i,&j)==0)	//type - recursion
//		if(isdiagz(array,m,n)==0)			//type - iteration
	    	printf("Diagonal not zero !\n");
    	else
    		printf("Diagonal is zero !\n");

no use at all
the out put is the same
please tell me whats wrong

Because unless you changed the i++ etc, you incremented the pointer, not what it was pointing at
To make the i you point at increment, you need (*i)++

again salem.........

1. this thought too had crossed my mind

Code:

   	    is = isdiagzero(a,m,n,(*i)++,(*j)++);		//is this parameter passing right i++,j++

it used to give me some compliation error like
"makes pointer from integer without a cast"
hence i felt that it was wrong assignment

2. let me tell you i tried this too

Code:

   	    is = isdiagzero(a,m,n,&((*i)++),&((*j)++));

which gives an error invalid value in unary'&'

3. also tried

Code:

   	    ti = (*i)++;
   	    tj = (*j)++;
   	    is = isdiagzero(a,m,n,&ti,&tj);

but the exception comes out

hence what else can it be?
i am not getting it.
please help

Code:

#include<stdio.h>

int isdiagzero(int a[][10], int m, int n, int *i, int *j)
{
    if (((*i) < m) && ((*j) < n)) {
        if (a[*i][*j] != 0)
            return 0;
        else {
            (*i)++;
            (*j)++;
            return isdiagzero(a, m, n, i, j);
        }
    }
    return 1;
}

int readmatrix(int a[][10], int *m, int *n)
{
    int i, j;
    printf("Enter the row and column ");
    scanf("%d%d", m, n);
    if ((*m == 0) || (*n == 0))
        return 0;
    else {
        printf("\nEnter the matrix\n");
        for (i = 0; i < *m; i++)
            for (j = 0; j < *n; j++)
                scanf("%d", &a[i][j]);
    }
    return 1;
}

void printmatrix(int a[][10], int m, int n)
{
    int i, j;
    printf("the matrix\n");
    for (i = 0; i < m; i++) {
        for (j = 0; j < n; j++)
            printf("%d ", a[i][j]);
        printf("\n");
    }
}

int isdiagz(int a[][10], int m, int n)
{
    int i, j;
    printf("inside isdiag\n");
    for (i = 0; i < m; i++)
        for (j = 0; j < n; j++)
            if ((i == j) && (a[i][j] != 0))
                return 0;
    return 1;
}

int main()
{
    int array[10][10];
    int m, n;
    int i = 0, j = 0;

    if (readmatrix(array, &m, &n) == 0)
        printf("ERROR\n");
    else {
        printmatrix(array, m, n);
        if (isdiagzero(array, m, n, &i, &j) == 0) //type - recursion
//      if(isdiagz(array,m,n)==0)           //type - iteration
            printf("Diagonal not zero !\n");
        else
            printf("Diagonal is zero !\n");
    }
    return 0;
}

i am still not getting it
please help me

Be more specific, I just tried it here

i tried the same code as such but still using the variable 'is'

Code:

int is = 0;
int isdiagzero(int a[][10],int m,int n,int *i,int *j)
{
    if(((*i)<m)&&((*j)<n))
    {
    if(a[*i][*j]!=0)
    	return 0;
   	else
   	{
   	    (*i)++;
   	    (*j)++;
   	    is = isdiagzero(a,m,n,i,j);		//is this parameter passing right i++,j++
    }
    }    
    return is;
}

now here it gives wrong output
i.e. for
0 1 1
1 0 1
1 1 0
the matrix
0 1 1
1 0 1
1 1 0
diagonal not zero !

does using variable 'is' cause a problem...
(since that is the only diifference in your code and mine)
if yes then what problem does it cause
please explain.

Originally Posted by aldajlo

(since that is the only diifference in your code and mine)

It's ok if you're stupid, but don't lie to us, because we are not:

Originally Posted by Salem

Code:

int isdiagzero(int a[][10], int m, int n, int *i, int *j)
{
    if (((*i) < m) && ((*j) < n)) {
        if (a[*i][*j] != 0)
            return 0;
        else {
            (*i)++;
            (*j)++;
            return isdiagzero(a, m, n, i, j);
        }
    }
    return 1;
}

Originally Posted by aldajlo

Code:

int is = 0;
int isdiagzero(int a[][10],int m,int n,int *i,int *j)
{
    if(((*i)<m)&&((*j)<n))
    {
    if(a[*i][*j]!=0)
    	return 0;
   	else
   	{
   	    (*i)++;
   	    (*j)++;
   	    is = isdiagzero(a,m,n,i,j);		//is this parameter passing right i++,j++
    }
    }    
    return is;
}

They are huh? The only thing different is a variable name? You want to try that again?

Quzah.

i didn't get it......
what did u mean Quzah.....

it's just using that variable thats making the output go wrong
and i don't get it . why?

please explain.......

> and i don't get it . why?
Because at some point, my function gets to return 1, and yours doesn't.