Function that return a structure that contains a pointer to array in C

Question

First few parts of code:

    typedef struct
    {
        double sr, med;
        int **t;
    }wynik;

    wynik calc(int *t[], int size)
    {
        int i, *niep = NULL, j = 0, k = 1, sum = 0;
        int *sorted = (int*)malloc(size*sizeof(int));
        wynik out;
        //coping, sorting
        for (i = 0; i < size; i++)
            sorted[i] = (*t)[i];
        qsort(sorted, size, sizeof (**t), cmp);
        out.t = &sorted;
...
    return out;
    }

then in main():

wynik get = calc(&tab, tab_size);

Using debugger I discovered that in calc() out.t points to an array, but in main() get.t points to some weird things. How to fix it?

Your sturcture doesn't contain an array. It contains a pointer. — juanchopanza
– juanchopanza, Commented Mar 10, 2015 at 14:27
I'm sorry but nothing in this code makes any sense. It would seem you are uncertain about how pointers and arrays work and have therefore obfuscated your program to the point where you no longer know what it does. On top of that you have memory leaks. This code is unsalvageable. — Lundin
– Lundin, Commented Mar 10, 2015 at 14:34

interjay · Accepted Answer · 2015-03-10 14:26:49Z

3

out.t contains the address of the local variable sorted. When the function returns, this address is no longer valid because the local variable went out of scope.

I see no reason here that out.t should be int** instead of int*. If you change it to be int* and just set its value with out.t = sorted, it should work correctly.

answered Mar 10, 2015 at 14:26

interjay

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3 Comments

Matt Ko Over a year ago

because he allocated the memory on the heap inside the function, that address should stay valid until it is freed, shouldn't it ?

interjay Over a year ago

@MattKo The address stored inside sorted (which was allocated by malloc) remains valid until freed. But the address of sorted itself (&sorted) becomes invalid when returning.

Matt Ko Over a year ago

You're right, I got a bit distracted, this code is not so easy to analyze. Of course, the address of sorted is on the stack, but its content on the heap.

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