C tridimensional array initialization

Question

I'd like to create a render table function in C, probably the worst language to do it.

I have a little problem initializing a table with is a bidimensional array of string, that make it a tridimensional array of char.

I am not able to initialize a tridimensionnal array that way:

char *table[][] = { 
    { "hello", "my", "friend" },
    { "hihi", "haha", "hoho"},
    NULL
};

But I get the error

test_table.c:8:11: error: array type has incomplete element type ‘char *[]’
     char *table[][] = {

Can C compiler compute the lengths of all the dimensions ?

I also have tried to declare it as table[][][] and all the possible variants with *....

The array dimensions except the first must be fixed (specified). Use char *table[][3], or perhaps char table[][3][7];. — Jonathan Leffler
– Jonathan Leffler, Commented Mar 31, 2016 at 20:18
Too bad the compiler can't compute it himself, it would not be that hard. Thank you for the help. — Nicolas Scotto Di Perto
– Nicolas Scotto Di Perto, Commented Mar 31, 2016 at 20:41

Jonathan Leffler · Accepted Answer · 2016-03-31 20:41:37Z

1

The array dimensions except the first must be fixed (specified). Use char *table[][3], or perhaps char table[][3][7]:

#include <stdio.h>

int main(void)
{
    char *table1[][3] =
    {
        { "hello", "my", "friend" },
        { "hihi", "haha", "hoho"},
    };

    char table2[][3][7] =
    {
        { "hello", "my", "friend" },
        { "hihi", "haha", "hoho"},
    };

    for (int i = 0; i < 2; i++)
        for (int j = 0; j < 3; j++)
            printf("[%d][%d] = [%s]\n", i, j, table1[i][j]);

    for (int i = 0; i < 2; i++)
        for (int j = 0; j < 3; j++)
            printf("[%d][%d] = [%s]\n", i, j, table2[i][j]);

    return 0;
}

Output:

[0][0] = [hello]
[0][1] = [my]
[0][2] = [friend]
[1][0] = [hihi]
[1][1] = [haha]
[1][2] = [hoho]
[0][0] = [hello]
[0][1] = [my]
[0][2] = [friend]
[1][0] = [hihi]
[1][1] = [haha]
[1][2] = [hoho]

There are other ways to do it too, such as using C99 'compound literals':

#include <stdio.h>

int main(void)
{
    char **table3[] =
    {
        (char *[]){ "hello", "my", "friend" },
        (char *[]){ "hihi", "haha", "hoho" },
        NULL,
    };

    for (int i = 0; i < 2; i++)
        for (int j = 0; j < 3; j++)
            printf("[%d][%d] = [%s]\n", i, j, table3[i][j]);

    return 0;
}

Output:

[0][0] = [hello]
[0][1] = [my]
[0][2] = [friend]
[1][0] = [hihi]
[1][1] = [haha]
[1][2] = [hoho]

answered Mar 31, 2016 at 20:41

Jonathan Leffler

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1 Comment

Nicolas Scotto Di Perto Over a year ago

I like the C99 'compounds literal' way, good trick ! Thanks !

Riad Baghbanli · Accepted Answer · 2016-03-31 20:22:14Z

0

See code below:

char *first[] = { "hello", "my", "friend" };
char *second[]  = { "hihi", "haha", "hoho" };
char **table[] = { first, second, NULL };

answered Mar 31, 2016 at 20:22

Riad Baghbanli

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