Parse in and return 2D array

Question

Trying to make a tic-tac-toe game and having trouble with the grid, I'm new to C and have looked everywhere but nothing seems to work.

int main(void) {
    char grid[GRID_HEIGHT][GRID_WIDTH];
    grid = make_grid((char **)grid);
    print_grid((char **)grid);
}

char ** make_grid(char **grid) {
    char grid[GRID_HEIGHT][GRID_WIDTH] = {
        { '\n', ' ', '1', ' ', '|', ' ', '2', ' ', '|', ' ', '3', ' ' } ,
        { '\n', '-', '-', '-', '-', '-', '-', '-', '-', '-', '-', '-' } ,
        { '\n', ' ', '4', ' ', '|', ' ', '5', ' ', '|', ' ', '6', ' ' } ,
        { '\n', '-', '-', '-', '-', '-', '-', '-', '-', '-', '-', '-' } ,
        { '\n', ' ', '7', ' ', '|', ' ', '8', ' ', '|', ' ', '9', ' ' }
    };
    return grid;
}

void print_grid(char **grid) {
    for (int row = 0; row < GRID_HEIGHT; row++) {
        for (int column = 0; column < GRID_WIDTH; column++) {
            printf("%c", grid[row][column]);
        }
    }
    printf("\n");
}

How would I parse the grid into a function and also return it without the program crashing?

Answer 1

You already declared the array grid in main.

char grid[GRID_HEIGHT][GRID_WIDTH];

So for example this function definition

char ** make_grid(char **grid) {
    char grid[GRID_HEIGHT][GRID_WIDTH] = {
        { '\n', ' ', '1', ' ', '|', ' ', '2', ' ', '|', ' ', '3', ' ' } ,
        { '\n', '-', '-', '-', '-', '-', '-', '-', '-', '-', '-', '-' } ,
        { '\n', ' ', '4', ' ', '|', ' ', '5', ' ', '|', ' ', '6', ' ' } ,
        { '\n', '-', '-', '-', '-', '-', '-', '-', '-', '-', '-', '-' } ,
        { '\n', ' ', '7', ' ', '|', ' ', '8', ' ', '|', ' ', '9', ' ' }
    };
    return grid;
}

does not make sense at least because the parameter is redeclared in the outer block scope of the function.

The function could look the following way

void make_grid( char ( *grid )[GRID_WIDTH] )
{
    static const char *rows[] = 
    {
        { "\n 1 | 2 | 3 " } ,
        { "\n-----------" } ,
        { "\n 4 | 5 | 6 " } ,
        { "\n-----------" } ,
        { "\n 7 | 8 | 9 " }
    };
    const size_t N = sizeof( rows ) / sizeof( *rows );

    for ( size_t i = 0; i < N; i++ )
    {
        strncpy( grid[i], rows[i], GRID_WIDTH );
    }
}

and called like

make_grid( grid );

The function print_grid can be defined like

void print_grid( char ( *grid )[GRID_WIDTH], size_t height ) 
{
    for ( size_t row = 0; row < height; row++) {
        for ( size_t column = 0; column < GRID_WIDTH; column++) {
            printf("%c", grid[row][column]);
        }
        putchar( '\n' );
    }
    putchar( '\n' );
}

and called like

print_grid( grid, GRID_HEIGHT );

Answer 2

Fundamental problems:

• You need to #include relevant headers.
• You need to declare functions before calling them.
• You cannot return arrays from functions in C. You can return pointers, but that might be the wrong solution. If you wish to copy an array, you must use memcpy .
• You cannot have two variables with the same name grid in the same scope.
• Pointer-to-pointers cannot point at arrays and they are not compatible with arrays. So you cannot use pointer-to-pointer here.

Fixed code:

#define GRID_HEIGHT 5
#define GRID_WIDTH 12

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

void make_grid (char grid[GRID_HEIGHT][GRID_WIDTH]);
void print_grid (char grid[GRID_HEIGHT][GRID_WIDTH]);

int main(void) {
    char grid[GRID_HEIGHT][GRID_WIDTH];
    make_grid(grid);
    print_grid(grid);
}

void make_grid (char grid[GRID_HEIGHT][GRID_WIDTH]) {
    const char GRID [GRID_HEIGHT][GRID_WIDTH] = {
        { '\n', ' ', '1', ' ', '|', ' ', '2', ' ', '|', ' ', '3', ' ' } ,
        { '\n', '-', '-', '-', '-', '-', '-', '-', '-', '-', '-', '-' } ,
        { '\n', ' ', '4', ' ', '|', ' ', '5', ' ', '|', ' ', '6', ' ' } ,
        { '\n', '-', '-', '-', '-', '-', '-', '-', '-', '-', '-', '-' } ,
        { '\n', ' ', '7', ' ', '|', ' ', '8', ' ', '|', ' ', '9', ' ' }
    };
    memcpy( grid, GRID, sizeof(char[GRID_HEIGHT][GRID_WIDTH]) );
}

void print_grid (char grid[GRID_HEIGHT][GRID_WIDTH]) {
    for (int row = 0; row < GRID_HEIGHT; row++) {
        for (int column = 0; column < GRID_WIDTH; column++) {
            printf("%c", grid[row][column]);
        }
    }
    printf("\n");
}

In this example, the parameter grid in each function is an array, which "decays" into a pointer to the first element. The first element of a 2D array is a 1D array, so you end up with an array pointer. If you don't want the compiler to "decay" the array you could write the array pointer syntax manually: char (*grid)[GRID_WIDTH] , but it is harder to read so I wouldn't recommend it.