How to fill a dynamic array in another function (c)

Question

I want to create a function, where I will scann elements (int) to an array and call this fuction in main, freeing memory also has to be in main.

int main () {
    int size = 0;
    int* arr = malloc (sizeof(int));
    char sign;
    int i = 0;
    do {
        scanf ("%d%c", &arr[i], &sign);
        i++;
    arr = realloc(arr, sizeof(int) * (i + 1));
    } while (sign != '\n');
    free(arr);
}

I 've tried to do this,

void initialize(int* arr) {
    char sign;
    int i = 0;
    do {
        scanf ("%d%c", &arr[i], &sign);
        i++;
        arr = realloc(arr, sizeof(int) * (i + 1));
    } while (sign != '\n');
}


int main () {
    int size = 0;
    int* arr = malloc (sizeof(int));
    
    initialize(arr);  
    free(arr);
  }

but I have an error

==140978== 1 errors in context 1 of 2:
==140978== Invalid free() / delete / delete[] / realloc()
==140978==    at 0x484B27F: free (in /usr/libexec/valgrind/vgpreload_memcheck-amd64-linux.so)
==140978==    by 0x1092A0: main (lol.c:21)
==140978==  Address 0x4a9a040 is 0 bytes inside a block of size 4 free'd
==140978==    at 0x484DCD3: realloc (in /usr/libexec/valgrind/vgpreload_memcheck-amd64-linux.so)
==140978==    by 0x109244: initialize (lol.c:10)
==140978==    by 0x109294: main (lol.c:19)
==140978==  Block was alloc'd at
==140978==    at 0x4848899: malloc (in /usr/libexec/valgrind/vgpreload_memcheck-amd64-linux.so)
==140978==    by 0x109284: main (lol.c:17)

Can I scann and fill a dynamic array in this way?

0___________ · Answer 1 · 2022-07-09T16:19:08.623

You need to pass pointer to pointer.
It is better to return size instead of relying on the sentinel values.
Readin input is not good. It will not work. A single read will terminate the loop.
Check the return value of scanf
Check the return value of realloc
Use objects not types in sizeof's

size_t initialize(int **arr) 
{
    size_t i = 0;
    int number;
    char line[100];
    while(fgets(line, sizeof(line), stdin) != NULL) 
    {
        if(sscanf(line, "%d", &number) == 1)
        {
            int *tmp;
            i++;
            tmp = realloc(*arr, sizeof(**arr) * i);
            if(tmp) 
            {
                *arr = tmp;
                (*arr)[i - 1] = number;
            }
            else 
            {
                /* error handling */
            }
        }
        else break;
    }
    return i;
}


int main (void) 
{
    size_t size;
    int *arr = NULL;
    
    size = initialize(&arr);  
    for(size_t i = 0; i < size; i++) printf("arr[%zu] = %d\n", i, arr[i]);
    free(arr);
}

https://godbolt.org/z/6zT47Kdxh

Thank you for your answer, but what is `line[100]` and will it be a problem, if my string will contain up to 10000 symbols, for example? — Катя Павліченко, Jul 10 '22 at 11:27

score 0 · Answer 2 · answered Jul 09 '22 at 16:02

0

Return the new pointer like realloc does it.

int* initialize(int* arr) {
    char sign;
    int i = 0;
    do {
        scanf ("%d%c", &arr[i], &sign);
        i++;
        arr = realloc(arr, sizeof(int) * (i + 1));
    } while (sign != '\n');
    return arr;
}


int main () {
    int size = 0;
    int* arr = malloc (sizeof(int));
    
    arr = initialize(arr);  
    free(arr);
  }

realloc is able to work with NULL. You can improve your code.

int* initialize(int* arr) {
    char sign;
    int i = 0;
    do {
        arr = realloc(arr, sizeof(int) * (i + 1));
        scanf ("%d%c", &arr[i], &sign);
        i++;
    } while (sign != '\n');
    return arr;
}

int main () {
    int size = 0;
    int* arr = initialize(NULL);  
    free(arr);
  }

answered Jul 09 '22 at 16:02

273K

29,503
10
41
64

Print entered values in main. Do you know how many numbers have been entered? – 0___________ Jul 09 '22 at 16:13
It seems you wanted to ask OP about this. – 273K Jul 09 '22 at 16:14
reading is invalid too. https://godbolt.org/z/afoGxPzfx guess why it does not read the numbers. DV-ted as you do posted code which does not work – 0___________ Jul 09 '22 at 16:16
No, I wanted to ask you. – 0___________ Jul 09 '22 at 16:18
Neither me want it, nor OP wants to print values. OP wants `free`. – 273K Jul 09 '22 at 16:20
Do you really think that he reads the values to free them? What about reading the values? Do you think that OP wants only one? – 0___________ Jul 09 '22 at 16:21
Pleas ask OP what they want. – 273K Jul 09 '22 at 16:22
A weakness with `int* initialize(int* arr)` is that the caller does not know how many `int` are allocated in the return pointer. – chux - Reinstate Monica Jul 09 '22 at 18:26

Viaceslavus · Answer 3 · 2022-07-09T16:17:14.333

When passing a pointer to a function and invoking realloc on it, you are freeing the memory that the pointer points to. So after the initialize function finishes its execution, you are calling free on an invalid pointer (it was already freed). What you should probably do is manipulating with a double pointer:

void initialize(int** arr) {
    char sign;
    int i = 0;
    do {
        scanf ("%d%c", &(*arr)[i], &sign);
        i++;
        *arr = realloc(*arr, sizeof(int) * (i + 1));
    } while (sign != '\n');
}

And invoking this function would look like this:

initialize(&arr);

This way you are writing the newly allocated address with realloc by unreferencing the pointer, so that after initialize function exits you will see the reallocated address.

Also it is also a good practice to check the value of a pointer after calling malloc and realloc. See proper usage of realloc

How to fill a dynamic array in another function (c)

3 Answers3