Question

I want to create a char** character array on the stack. Currently, I’m using this, but I wonder if there is a nicer way:

char* buf[4];
char temp0[1024], temp1[1024], temp2[1024], temp3[1024];
buf[0] = temp0;
buf[1] = temp1;
buf[2] = temp2;
buf[3] = temp3;

EDIT: To be more clear, I can not simply use char buf[4][1024]. A function expecting an array of char pointers would crash, as it’s a fundamentally different data type. This is how I would create the array on the heap:

char** buf = malloc(sizeof(char*) * 4);
for(int i = 0; i < 4; i++)
{
    buf[i] = malloc(sizeof(char) * 1024);
}

Answer 1

The solutions posted so far are both OK for 4 elements; for 10 they’re clunky, and for 100 not going to fly. I think this can scale better:

enum { MAX_ROWS = 10, ROW_SIZE = 1024 };
char bigbuffer[MAX_ROWS][ROW_SIZE];
char *buf[MAX_ROWS];

for (int i = 0; i < MAX_ROWS; i++)
    buf[i] = bigbuffer[i];

...and off you go...

With C99 or later, you can parameterize the array size by using VLAs (variable length arrays):

void buffer_creation(int rows, int cols)
{
    char bigbuffer[rows][cols];
    char *buf[rows];

    for (int i = 0; i < rows; i++)
        buf[i] = bigbuffer[i];

    ...and off you go...
}

If the size gets too large, you can use malloc() instead, of course, but you have to make sure you free the space too:

void buffer_creation(int rows, int cols)
{
    char *buf[rows];  // Probably won't stress the stack ever
    char *bigbuffer = malloc(rows * cols);
    if (bigbuffer != 0)
    {    
        for (int i = 0; i < rows; i++)
            buf[i] = &bigbuffer[i * cols];

        ...and off you go...
        free(bigbuffer);
    }
}

Obviously, you could allocate the buf array too if you really want to — I’m leaving it as an exercise for the reader.