Stuck while printing a 2D array's border with the char "*"

Question

Basically what I need to do is print a border of "*" using the 2D array in C. However I managed to print somewhat one part of the border but now I am stuck in printing the other side. I cannot seem to figure out the right loop to do in this case.

I am new to C coding or any type of coding.

-Has to be a 2D array

-Contains spaces in between

-Only prints the border

eg:

***********
*         *
*         *
***********

I have attached my code so far and my result of that code.

PS: I have used malloc to allocate the 2Darray

for(kk=0;kk<10; kk++)
    {
        for(ll=0;ll<20; ll++)
        {
            if (kk == 0)
                {
                    basicMap[kk][ll] = '*';
                    printf("%c", basicMap[kk][ll]);
                }
            else if(kk == 9)
            {
                basicMap[kk][ll] = '*';
                printf("%c", basicMap[kk][ll]);
            }
            else if (ll ==0)
            {
                basicMap[kk][ll] = '*';
                printf("%c", basicMap[kk][ll]);
            }
            else if(ll == 19)
            {
                basicMap[kk][ll] = '*';
                printf("%c", basicMap[kk][ll]);
            }

        }
        printf("\n");
    }

result of above code

Answer 1

You never print the spaces. Simplifying the code might make that clearer. You've got a lot of redundant code that can be eliminated. Something like:

for( kk = 0; kk < 10; kk += 1 ){
    for( ll = 0; ll < 20; ll++ ){
        if( kk == 0 || kk == 9 || ll == 19 || ll == 0 ){
            basicMap[kk][ll] = '*';
        } else {
            basicMap[kk][ll] = ' '; /* Perhaps redundant, if previously assigned */
        }
        putchar(basicMap[kk][ll]);
    }
    putchar('\n');
}

Answer 2

If I have understood correctly you are trying to do something like the following shown in the demonstrative program below.

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

int main(void) 
{
    size_t m = 4;
    size_t n = 11;
    
    char **a = malloc( m * sizeof( char * ) );
    
    for ( size_t i = 0; i < m; i++ )
    {
        a[i] = malloc( n * sizeof( char ) );
        
        if ( i % ( m - 1 ) == 0 )
        {
            memset( a[i], '*', n ); 
        }
        else
        {
            memset( a[i] + 1, ' ', n - 2 );
            a[i][0] = a[i][n-1] = '*';
        }
    }
    
    for ( size_t i = 0; i < m; i++ )
    {
        printf( "%.*s", ( int )n, a[i] );
        putchar( '\n' );
    }
    
    for ( size_t i = 0; i < m; i++ )
    {
        free( a[i] );
    }
    
    free( a );
    
    return 0;
}

The program output is

***********
*         *
*         *
***********