Thread: Variable length 2d array

Hi guys,

this is a somewhat philosophical question.

I was searching for a way to minimize memory usage in an embedded system and come up with this solution:

Code:

#include <stdio.h>
int main()
{
    char* t[]={"abc","d","ef"}; // Variable length 2d array
    int i;
    for (i=0; i<9; i++)
    {
        printf("%d,",t[0][i]); // Shows bytes in array
    }
    printf("\n");
    printf("%s\n",t[0]); // Shows first string
    printf("%s\n",t[1]); // Shows second string
    printf("%s\n",t[2]); // Shows third string
    printf("\n");
    return 0;
}

Here is the output for this code:

Code:

sh-4.3$ gcc -o main *.c                                              
sh-4.3$ main                                               
97,98,99,0,100,0,101,102,0,
abc
d
ef
sh-4.3$

As you can see, the compiler does not created a 3x3 array, but an array with 3 elements, where the first is 4 bytes long, the second is 2 bytes long, and the third is 3 bytes long. Every string has a zero to finalize it.

Despite this strange format, a reference to a given index works fine. You can get the second element with t[1].

Here is my question: I always thought C finds an element in a 2d array by multiplying the given references like this:

Code:

If I have t[5][5],
t[3][4] == t[0][3*5+4]

But how C finds an element in this variable length 2d array that I described here?

O_o

You don't have an array with two dimensions. You have an array of pointers.

Your code may appear to work, but the application doing what you expect is just a coincidence of where the compiler stored the strings.

The syntax is also not strange.

Code:

int x = 0, y = 1, z = 2;
int s[] = {x, y, z};

Code:

const char * x = "abc", * y = "d", * z = "ef";
const char * s[] = {x, y, z};

Soma