How to access a 2d array inside a struct using only pointers

Question

Trying to understand pointers as a beginner in C- I've got this struct:

    typedef struct {
        int size;           // dimension of array
        int **arr; // pointer to heap allocated array
    } someStruct;

So I use malloc to generate this struct, and an array, and initialize all the values to zero-

someStruct *m = (someStruct*)malloc(sizeof(someStruct));
m->size = n;
m->arr = (int**)malloc(n * sizeof(int));
// initialize array
for (int i = 0; i < n; i++) {
    *(m->arr + i) = (int*)malloc(n * sizeof(int));
    // set value to 0
    for (int j = 0; j < n; j++) {
        *(*(m->arr + i) + j) = 0;
    }
}

After this I basically continue to access the array in later stages using the same kind of pointer logic-

for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
        int num =  *(*(m->arr + i) + j);
        printf("num: %d\n", num);
    }
}

Here's the problem- when I try to use this method of access, I'm clearly not getting the right answer- my print output look like this:

num: -2043774080
num: 22031
num: 0
num: 0
...
num: 0
num: 0

Here's the really weird part- this seeming bug of the 'weird' random numbers only comes when I'm creating and accessing an array of size 5-

I've come to believe that the whole

*(*(m->arr + i) + j)

method of access must be wrong- any help on this would be really useful. Thanks in advance, I apologize if this was already answered, my searching was unable to find it.

malloc(n * sizeof(int)) allocates storage for n integer values, but for the outer level you are using this as if it's storage for n pointer values. If sizeof(int) < sizeof(int*) on your system, you're in for a bad time. Now, please stop using pointer dereferencing for array-like access. *(*(m->arr + i) + j); is equivalent to m->arr[i][j] -- one of these is readable, and one is not. — paddy
– paddy, Commented Feb 23, 2022 at 1:14
@paddy so would the problem be in the access code or when I typecast to int* during the assignment? — Pranav Nair
– Pranav Nair, Commented Feb 23, 2022 at 1:35
m->arr = (int**)malloc(n * sizeof(int)); is a problem. Avoid guessing the type: Use m->arr = malloc(sizeof m->arr[0] * n);. — chux
– chux, Commented Feb 23, 2022 at 1:46

SGeorgiades · Accepted Answer · 2022-02-23 01:30:41Z

1

You should give complete code, but I think I was able to figure out your intent. You have one glaring problem, and many style issues. Here is what I think your code should look like:

typedef struct {
    int size;           // dimension of array
    int **arr; // pointer to heap allocated array
} MagicSquare;

  :
  :

// no need to dynamically allocate this, it is small
MagicSquare m;
m.size = n;
m.arr = malloc(n * sizeof(int*));  // note it is sizeof(int*), not (int)

// initialize array
for (int i = 0; i < n; i++) {
    m.arr[i] = malloc(n * sizeof(int));
    // set value to 0
    for (int j = 0; j < n; j++) {
        m.arr[i][j] = 0;
    }
}

  :
  :

for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
        printf("num: %d\n", m.arr[i][j]);
    }
}

Note that if you want to initialize the allocated memory to zero, you should just use calloc, which does this initialization for you:

// initialize array
for (int i = 0; i < n; i++) {
    m.arr[i] = calloc(n,sizeof(int));
}

answered Feb 23, 2022 at 1:30

SGeorgiades

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4 Comments

chux Over a year ago

m.arr = malloc(n * sizeof m.arr[0]); avoids type mis-match better than trying to synchronize the pointer and its type.

SGeorgiades Over a year ago

True, @ReinstateMonica... or m.arr = malloc(n * sizeof *m.arr); Similarly, we could use m.arr[i] = calloc(n,sizeof *m.arr[i]);

Pranav Nair Over a year ago

Hey, thanks for the answer, the fix works- I'm just trying to understand this- a lot of online tutorials seem to like to use this whole typecast thing of calling m->arr = (int*)malloc(n * sizeof(int)); is this method of doing it completely invalid? I also tried to format the question better, apologize for that mess.

paddy Over a year ago

In C, it's advisable to omit the cast. The typecast is required for the code to be compiled with a C++ compiler. Of course, if you're using C++ then using malloc at all is not generally recommended. Unfortunately the quality of "online tutorials" is quite a mixed bag.

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