Why is realloc() introducing some random values into a dynamic array when called multiple times?

Question

I am trying to fill a dynamically generated array using a for loop and make it bigger whenever needed using realloc(). However, when realloc is called several times on the memory, it replaces some values, especially the first and last few with seemingly random numbers.

This is the code I tested the said with:

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

void fill(int * array, int size, int number_of_values) {
    for (int i = 0; i < number_of_values; i++) {
        if (i == size) {
            size *= 2;
            array = realloc(array, sizeof(int) * size);
        }

        array[i] = 100;
    }
}

int main() {
    int *array;
    int number_of_values = 21;
    int size = 5;

    array = malloc(sizeof(int) * size);

    fill(array, size, number_of_values);

    for (int i = 0; i < number_of_values; i++) {
        printf("%i ", array[i]);
    }
}

I tried reimplementing the code several times and looked at the array directly after assigning 100 to it, but all the values are assigned correctly at first and then change somehow. I also observed, that said behavior only seems to happen when there are more than 20 items assigned to the array in the loop.

The output of the program normally looks something like this:

10425976 10436440 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 134217736

Whereas when I set number_of_values to 20 and therefore only assign 20 values, the output looks like the following:

100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100 100

I also looked through some references of the realloc function, but as far as I'm concerned, I used it correctly.

You discard the pointer returned by realloc. (You change fill's array, but not main's.) You are using a pointer that's no longer valid. — ikegami
– ikegami, Commented Apr 28, 2020 at 19:29
Your print loop prints values from the array created by the very first call to malloc, not by realloc. — gudok
– gudok, Commented Apr 28, 2020 at 19:31
array = realloc(array,...) is a memory leak in case the realloc fails. You should always assign the return value of realloc to another pointer to check it. — Werner Henze
– Werner Henze, Commented Apr 28, 2020 at 19:31
@ikegami Doesn't array = realloc(array, sizeof(int) * size); set array to the new pointer? — Vntto
– Vntto, Commented Apr 28, 2020 at 19:31
Yes, that changes fill's array. Not main's, though. There are two variables named array, and you are only changing one of them. — ikegami
– ikegami, Commented Apr 28, 2020 at 19:31

john · Accepted Answer · 2020-04-28 19:36:59Z

1

The problem is that your function must return the newly allocated array.

int* fill(int * array, int size, int number_of_values) {
    ...
    return array;
}

int main() {
    ...
    array = fill(array, size, number_of_values);
    ,,,
}

Beginners generally get this wrong for a couple of reasons. They might think that because the two variables called array in main and in find both have the same name they are the same variable. This is not true.

The other reason is that sometimes beginners think that there is something special about pointers. If you change a pointer inside a function somehow it also changes the corresponding pointer in the calling function. I think this is a misunderstanding based on the fact that you can use a pointer to change the value being pointed at, but the pointer itself has the same rules as any other type of variable.

edited Apr 28, 2020 at 19:36

answered Apr 28, 2020 at 19:31

john

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4 Comments

Vntto Over a year ago

Thanks, that is exactly what I thought. Several other people who are also learning C had the same issue.

john Over a year ago

@Vntto Which one did you think, the first one or the second one?

Vntto Over a year ago

The second one, I didn't realize, that I am not changing the being pointed at, but the pointer itself

john Over a year ago

@Vntto Yes, it's a common misunderstanding.

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