Write a C program to create identity matrix or unit matrix

[linda7]

Hi

I'm writing a program that creates an identity matrix. when I run this program it gives me an error message.

Any help please !

[linda7]

Hi

I'm writing a program that creates an identity matrix. when I run this program it gives me an error message.

Any help please !

#include<stdio.h>
#include<stdlib.h>
int main () {
	
	
	int i,j,m,n;
	int a;
	
	printf(" Please enter the order of the matrix m x n \n");
	scanf("%d%d",&m,&n);

	a = (int *) malloc(m*n*sizeof(int *)); /* Row pointers */
 
    if (m==n){
	for(i=0;i<m;i++){
		
		for(j=0;j<m;j++) {
		
			if (i==j) {
			a[i][j]=1; 
	        printf("%d ",a[i][j]);   }
			else{
			a[i][j]=0;
			printf("%d ",a[i][j]);   }	
		}
	}
	printf("Matrix A is\n");
    for(i=0;i<m;i++)
    {
        for(j=0;j<m;j++)
        printf("%d\t",a[i][j]);
        printf("\n");
    }
    
	}// end if
	
	return 0;
}

[stdq]

What are the lines that you're getting errors at? You need a pointer to pointer to hold the address of the first element of the matrix. Assuming you use a, this variable should be declared as:

int **a;

Allocate the rows with the following statement:

a = ( int ** )malloc( m * sizeof( int * ) );

Then, use a for statement to allocate memory for the columns of each row:

for ( i = 0 ; i < m ; ++i )
{
    *( a + i ) = ( int * )malloc( n * sizeof( int ) );
}

Now, loop through the whole matrix and assign 1 or 0 to each element:

for ( i = 0 ; i < m ; ++i )
{
    for ( j = 0 ; j < n ; ++j )
    {
        if ( i != j )
        {
            a[ i ][ j ] = 0;
        }
        else
        {
            a[ i ][ j ] = 1;
        }
    }
}

Then, print the matrix to check that everything is correct:

printf( "The matrix is:\n\n" );

for ( i = 0 ; i < m ; ++i )
{
    for ( j = 0 ; j < n ; ++j )
    {
        printf( "%3d", a[ i ][ j ] );
    }

    putchar( '\n' );
}

[linda7]

error message: subscripted value is neither array nor pointer ( lines 20, 21, 23, 24, 31).

Do I have to use double pointer ( int **a ), I mean is there another way to do this ??
And why it gives these errors !!

[stdq]

You're probably getting these errors because a is simply an integer, and not a pointer to pointer. I'm not really sure you can do the following, which also ends up using pointer indirectly:

int a[ m ][ n ];

If this is correct, the memory is allocated at compile time (I think). Other than that, I can't think of other solution. And like I said, the above line might be incorrect - and maybe that depends on the compiler you're using.

[linda7]

Yes it gives error: "incompatib le type", it is fine I will consider the above solution.

Thank you so much

[stdq]

You're welcome. And I forgot: check whether each allocation returns NULL. Also, free all the memory allocated. I think you could do that like this:

free( a );
a = NULL;

for ( i = 0 ; i < m ; ++i )
{
    free( *( a + i ) );
    *( a + i ) = NULL;
}

Someone please correct me if I'm wrong. Here is a complete program, for a matrix that is 4x4, although it might not be perfect:

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

int main( void )
{
    int **a;
    int i;
    int j;
    int m = 4;
    int n = 4;
    
    a = ( int ** )malloc( m * sizeof( int * ) );
    
    if ( a == NULL )
    {
        printf( "Memory error...\n" );
        
        /* To indicate failure: */
        return 1;
    }
    
    for ( i = 0 ; i < m ; ++i )
    {
        *( a + i ) = ( int * )malloc( n * sizeof( int ) );
        
        if ( *( a + i ) == NULL )
        {
            printf( "Memory error...\n" );
        
            /* To indicate failure: */
            return 1;
        }
    }

    for ( i = 0 ; i < m ; ++i )
    {
        for ( j = 0 ; j < n ; ++j )
        {
            if ( i != j )
            {
                a[ i ][ j ] = 0;
            }
            else
            {
                a[ i ][ j ] = 1;
            }
        }
    }

    printf( "The matrix is:\n\n" );
     
    for ( i = 0 ; i < m ; ++i )
    {
        for ( j = 0 ; j < n ; ++j )
        {
            printf( "%3d", a[ i ][ j ] );
        }
     
        putchar( '\n' );
    }
    
    /******** I'm not sure my deallocation is correct: /*********/
    free( a );
    a = NULL;
    for ( i = 0 ; i < m ; ++i )
    {
        free( *( a + i ) );
        *( a + i ) = NULL;
    }
    
    /* Successful program termination: */
    return 0;
}

[linda7]

But I don't need a dynamic array, I just need to make it optional for the user to enter any size of array and then this size become static ( I mean the array become static ), so I don't need to free the memory because no extra memory is available, am I right ??

[stdq]

If you use malloc, the memory should be freed. At least I think so.