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?

Also, as soon as you return from `make_grid`, `grid` is gone, you can't reference it. — Federico klez Culloca, Dec 03 '19 at 10:22
Note that you can't pass a 2D array to `void print_grid(char **grid)`, it should be `void print_grid(char grid[][GRID_WIDTH])`. — Weather Vane, Dec 03 '19 at 10:29
Changing void print_grid(char **grid) to print_grid(char grid[][GRID_WIDTH]) seems to have worked, thanks — aLongBoat, Dec 03 '19 at 10:38
With what values? Please post the full code so its details don't have to be extracted bit by bit, even several questions about the same values. — Weather Vane, Dec 03 '19 at 10:39
Your return type is wrong, ** is a pointer to a pointer, you just need to return char*. However you need to make your array static as right now its local and because its local it will only exist on the stack for the life of the function call. — SPlatten, Dec 03 '19 at 10:42
Why was this closed as duplicate? The dupe only addresses one of many problems with this code. Re-opening. — Lundin, Dec 03 '19 at 11:36

score 1 · Answer 1 · edited Dec 03 '19 at 10:54

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 );

Do not use `strncpy`, it is a dangerous function that causes bugs such as `strncpy( grid[i], rows[i], GRID_WIDTH );`. The resulting arrays are not null terminated because `GRID_WIDTH` doesn't contain space for the null terminator, as the code is written. Use `memcpy` instead. — Lundin, Dec 03 '19 at 11:45
@Lundin It is not a bug. It is a design solution when the character array does not contain the terminating zero. Why did you decide that the element of grid shall be zero=terminated?! — Vlad from Moscow, Dec 03 '19 at 11:48
Because `strncpy` is just the lazy version of `memcpy`. You know the size, you know that the strings aren't null terminated, so there's no point to call a function slowly searching for the null termination that isn't there. — Lundin, Dec 03 '19 at 12:05
@Lundin I think that in this case it is better to use strncpy instead of memcpy because there is nothing can be said about the initializers. — Vlad from Moscow, Dec 03 '19 at 12:21

score 0 · Answer 2 · answered Dec 03 '19 at 11:41

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.

Parse in and return 2D array

2 Answers2