Thread: Array with pointer to Structarray

Hi,

Code:

#include <stdio.h>
#include <stdlib.h>

#define LENGTH(x) (sizeof(x)/sizeof(x[0]))

struct foo {
    int a;
    char* b;
};

struct foo bar[] = {
    {1, "abc"}, {2, "def"}, {3, "ghj"}
};

struct foo baz[] = {
    {1, "xyz"}
};

struct foo* array[] = {
    bar, baz
};

int main(void) {
    int i, j;

    printf("%s\n", array[0][2].b);
    printf("%d %d\n", (int)LENGTH(bar), (int)LENGTH(array[0]));
    
    for(i = 0; i < LENGTH(array); i++)
        for(j = 0; j < LENGTH(array[j]); j++)
            printf("%d: %s\n", array[i][j].a, array[i][j].b);

    return EXIT_SUCCESS;
}

the bold macro call returns 0. I guess the complier knows the struct-array size only at compile time so my dynamic solution wont work. Any ideas for a solution without saving the array size?

Regards,
lwi

Of course you are right. I just figured out myself I cannot get the array-size when I only have a pointer to the array.

I guess the best solution would be to NUL-terminate my struct-array, however I dont know exactly how to do it, probably a new struct (

Code:

struct { struct foo[] array; char nul; }

)?

Regards,
lwi

Thanks, works perfectly.

Solution:

Code:

#include <stdio.h>
#include <stdlib.h>

#define LENGTH(x) (sizeof(x)/sizeof(x[0]))

struct foo {
    int a;
    char* b;
};

struct foo bar[] = {
    {1, "abc"}, {2, "def"}, {3, "ghj"}, {0, NULL}
};

struct foo baz[] = {
    {1, "xyz"}, {0, NULL}
};

struct foo* array[] = {
    bar, baz
};

int main(void) {
    int i, j;

    printf("%s\n", array[0][2].b);
    printf("%d %d\n", (int)LENGTH(bar), (int)LENGTH(array[0]));
    
    for(i = 0; i < LENGTH(array); i++)
        for(j = 0; !(array[i][j].a == 0 && array[i][j].b == NULL); j++)
            printf("%d: %s\n", array[i][j].a, array[i][j].b);

    return EXIT_SUCCESS;
}