I need help for bubble sort in c

Question

How can I display the input user of this code. I am using bubble sort but I cannot display the input.

Here's my code:

    #include <stdio.h>

    int main(void) {
    int array[100];

    int count;

    for(int i = 0; i < 100; i++) {
        printf("Enter number %d: ", i + 1);
        scanf("%d", &array[i]);
        if(array[i] == 0) break;
    }

    int size = sizeof(array) / sizeof(array[0]);
    for(int i = 0; i < size - 1; ++i) {
        int swapped = 0;
        for(int ctr = 0; ctr < size - i - 1; ++ctr) {
            if(array[ctr] > array[ctr + 1]) {
                int temp = array[ctr];
                array[ctr] = array[ctr + 1];
                array[ctr + 1] = temp;

                swapped = 1;
            }
        }
        if(swapped == 0) {
            break;
        }
    }
    
    printf("[");
    for(int i = 0; i < size; i++) {
        if(i == size - 1) {
            printf("%d", array[i]);
        }
        else {
            printf("%d, ", array[i]);
        }
    }
    printf("]");
        
    return 0;
    }

Here's the output:

Enter·number·1:·1
Enter·number·2:·2
Enter·number·3:·3
Enter·number·4:·4
Enter·number·5:·5
Enter·number·6:·0
[1,·2,·3,·4,·5]

I cannot display the 1 2 3 4 5.... How can I display?

Answer 1

The calculation of the size doesn't work like you did it. sizeof(array) will give you the total amount of of bytes of the array (of all 100 spaces). So your size will allway be 100.

Instead of calculating the size it's easyer to count the inputs like this:

int size=0;

for(int i = 0; i < 100; i++) {
    printf("Enter number %d: ", i + 1);
    scanf("%d", &array[i]);
    if(array[i] == 0) break;
    size++;
}