different output with different version of gcc, when code dynamic array using c

Question

I am code a program about dynamic array in c language, the code is :

#include <stdio.h>

struct Vector {
    int size;
    int capacity;
    int *arr;
};

void add(struct Vector *Arr, int data) {

    if (Arr->size == Arr->capacity) {
        Arr->capacity *= 2;
        int arr[Arr->capacity];

        //array copy
        for (int i = 0; i < Arr->size; i++) {
            arr[i] = Arr->arr[i];
        }

        Arr->arr = arr;
    }

    int size = Arr->size;
    Arr->arr[size] = data;
    Arr->size++;

}

void display(struct Vector *Arr) {
    for (int i = 0; i < Arr->size; i++) {
        printf("%d ", Arr->arr[i]);
    }
    printf("\n");
}

int main() {
    int arr[10];
    struct Vector
            array = {0, 10, arr};

    //fill the array
    for (int i = 0; i < 10; i++) {
        add(&array, i);
    }
    display(&array);
    //more element than the init size
    add(&array, 10);
    display(&array);    //where the error happened

    return 0;
}

When the array growth, it has defferent output like below:

using dev-cpp with gcc 4.9:

using vs code with gcc8.2

using the online c compiler:

And the last one is my expectation.

^ Which is why you should not be tagging with a language you aren't programming to. — StoryTeller - Unslander Monica
– StoryTeller - Unslander Monica, Commented Apr 2, 2019 at 6:41
More or less a duplicate of this: stackoverflow.com/q/4824342/898348 and this: stackoverflow.com/q/4570366/898348 — Jabberwocky
– Jabberwocky, Commented Apr 2, 2019 at 6:47
Thanks, I've cheaked them which are the same problem as mine. — Flux
– Flux, Commented Apr 2, 2019 at 7:23

Blaze · Accepted Answer · 2019-04-02 06:49:18Z

5

The problem is that you have undefined behavior, so anything might happen. It can manifest in different ways on different machines or compilers. Consider this:

if (Arr->size == Arr->capacity) {
        Arr->capacity *= 2;
        int arr[Arr->capacity];
        ...
        Arr->arr = arr; // Arr->arr points to a local variable!

Here you're making a new array and then you are assigning its address to the vector. However, when that function finishes, that memory becomes invalid. Instead, replace it with this:

int *arr = malloc(sizeof(int) * Arr->capacity);

And you get the following output:

0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9 10

And don't forget to free it when you're done. To get that to work properly, I would recommend changing the int arr[10]; to int arr = malloc(10*sizeof(int)); so the array never is on the stack, then put a free(Arr->arr); before the Arr->arr = arr; as well as a free(array.arr); at the end of the program.

edited Apr 2, 2019 at 6:49

answered Apr 2, 2019 at 6:43

Blaze

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4 Comments

Richard Hodges Over a year ago

he has an additional problem. The initial array storage is initialised with a literal. You can't free the address of that.

Blaze Over a year ago

@RichardHodges that's right, which is why I suggested replacing int arr[10]; with int arr = malloc(10*sizeof(int)); so all of it can be freed.

Richard Hodges Over a year ago

Ah yes, didn't see that part. My apologies.

Flux Over a year ago

Got it! It is easily understood, nice answer.

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