Question

Trying to teach myself how to pass an array from int main() to a function and I just don’t understand it.

Here is my code.

#include <stdio.h>

void printboard( char *B ) {
 /*Purpose: to print out the tic tac toe board B
  */
    int i,j;

    printf("\n");
    for (i=0;i<3;i++) {
        for (j=0;j<3;j++) {
            printf( "      %c ",B[i][j]);
        }
        printf("\n\n");
    }
}

int main( void ) {
    char B[3][3] = { '-','-','-',
                     '-','-','-',
                     '-','-','-' };

    printboard(B);
}

I get this error:

test.c: In function 'printboard':
test.c:12: error: subscripted value is neither array nor pointer
test.c: In function 'main':
test.c:25: warning: passing argument 1 of 'printboard' from incompatible pointer type

Just need to get an understand of how pointers work and are passed so I can go ahead with my work.

Answer 1

Two dimensional array decays to a pointer to one dimensional array (i.e., (*)[] )

So, change from –

void printboard( char *B ) 
{
   // ....
}

to

void printboard( char B[][3] )
                 // Compiler is not bothered about first index because it 
                 // converts to (*)[3]. So, is the reason it left empty.
{
    // ...
}

---

Also, two dimensional array initialization should be done this way –

char B[3][3] = { {'-','-','-'},
                 {'-','-','-'},
                 {'-','-','-'}
               };

Check Results