动态数组使用malloc和realloc？

Question

I'm trying to collect input integers one by one. My array starts with size 1 and I want to expand it by 1, with every input collected (is this a smart thing to do?)

Anyway, this is the code I managed to come up with, but it's not working as intended.

After this process, sizeof(array) always returns 8, which I assume means that the array is only being resized once. (sizeof(int) is 4 bits)

Trying to output the array results in multiple instances of the first input variable.

OUTPUT CODE

for(int s=0;s<sizeof(array)/sizeof(int);s++){
    printf("%i\n",array[i]);
}

ORIGINAL CODE:

    int i;
    int size = 1;
    int *array = malloc(size * sizeof(int));
    int position = 0;

    do{
        i = getchar() - 48;
        f (i != -16 && i != -38 && i != 0) {
            array[position] = i;

            position++;
            size++;
            *array = realloc(array, size * sizeof(int));
        }
    } while (i != 0);

UPDATED STILL NOT WORKING CODE

    int i;
    int size = 1;
    int *array = malloc(size * sizeof(int));
    int position = 0;

    do{
        i = getchar() - 48;
        f (i != -16 && i != -38 && i != 0) {
            array[position] = i;

            position++;
            size++;
            array = realloc(array, size * sizeof(int));
        }
    } while (i != 0);

Answer 1

array = realloc(...)

not *array . Per the realloc docs , realloc returns the pointer, which you can store directly in your array pointer.

Edit One thing that will make your life easier: use char constants instead of raw numbers. Eg,

i = getchar();
if(i != ' ' && i != '\n' && i != '0') {
 /*    48-16       48-38        48-0      right? */
    array[position] = i - '0';   /* '0' = 48 */

Answer 2

One thing that jumps out at me: inside your loop, this line:

*array = realloc(array, size * sizeof(int));

should instead be:

array = realloc(array, size * sizeof(int));

In the original version, you were sticking the result of realloc in the first element of the array by dereferencing the pointer first. Without the asterisk, you're reassigning the array itself.

Answer 3

(With some copy-paste from my comment:) sizeof(array) returns 8 because it equals sizeof(int*) ( array is type int* ) which is 8 (you're probably compiling as 64-bit). sizeof doesn't work how you think for pointers to arrays.

Similarly, your output code is wrong, for the same reason. You only print the first two elements because sizeof(array)/sizeof(int) will always be 8/4=2 . It should be

for(int s=0;s<size;s++){
    printf("%i\n",array[s]);
}

(note also changed index variable i to s ) where size is the variable from your other code chunk(s). You cannot find the length of the array from sizeof if it's dynamically allocated with pointers; that's impossible. Your code must "remember" the size of your array.