How to implement a 2-dimensional array of struct in C

Question

I'm currently trying to understand how to implement a 2-dimensional array of struct in C. My code is crashing all the time and I'm really about to let it end like all my approaches getting firm to C: garbage. This is what I got:

typedef struct {
    int i;
} test;

test* t[20][20];
*t = (test*) malloc(sizeof(test) * 20 * 20);

My glorious error:

error: incompatible types when assigning to type ‘struct test *[20]’ from type ‘struct test *’

Do I have to allocate the memory seperately for every 2nd dimension? I'm getting nuts. It should be so simple. One day I will build a time-machine and magnetize some c-compiler-floppies...

Answer 1

This should be enough:

typedef struct {
    int i;
} test;

test t[20][20];

That will declare a 2-dimensional array of test of size 20 x 20. There's no need to use malloc.

If you want to dynamically allocate your array you can do this:

// in a function of course
test **t = (test **)malloc(20 * sizeof(test *));
for (i = 0; i < 20; ++i)
    t[i] = (test *)malloc(20 * sizeof(test));

Answer 2

test **t;

t = (test **)malloc(sizeof(test *) * 20);
for (i = 0; i < 20; i++) {
   t[i] = (test *)malloc(sizeof(test) * 20);
}

Answer 3

Other answers show how to fix it but they don't explain why. As the compiler hinted, the type of t in your original example is actually test *[20] which is why your cast to test * wasn't enough.

In C, the name of an array T of dimension N is actually of type *T[dim0][dim1]...[dimN-1]. Fun.