What makes a program not follow the loop condition?

Question

Just a basic program, written by a newbie. Supposed to read in values into a dynamic array, then print them out in reverse order. You can't input the size of array beforehand, the array's size will adjust as long as inputs aren't terminating characters.

I think I wrote it okay, but there are errors about dereferencing pointers and when I run it in VS, it won't even register inputs. When attempted in other compiler, it does register the input, but doesn't terminate with "-1".

Thinking about it, looking it up, I don't notice my mistake, hope you will help me.

Edit: thanks for pointing out the semicolon after while, but now it's a "Heap Corruption Error" after inputting 2 or 3 inputs. What went wrong?

int i=0;
    int *p, *a;
    int n=1;
    p = (int*)malloc(n * sizeof(int));
    printf("Enter integers here, and input -1 when done:\n");
    while (p[i] != -1);
    {
        scanf_s("%d", &p[i]);
        n = i + 1;
        a= (int*)malloc(n * sizeof(int));
        for (int j = 0; j < n; ++j)
        {
            a[j] = p[j];
        }
        free(p);
        p = NULL;
        p= (int*)malloc(n * sizeof(int));
        for (int k = 0; k < n; ++k)
        {
            p[k] = a[k];
        }
        free(a);
        a = NULL;
        ++i;
    }
    --i;
    if (i <= 0)
    {
        printf("%d", p[i]);
        --i;
    }
    free(p);
    p = NULL;

`while (p[i] != -1);` `p[0]` isn't initialized, so this is undefined behavior. — Blaze, Oct 25 '19 at 07:48
In passing, I recommend that you read and understand [the question on why not to cast the return value of `malloc()` and family in C](/q/605845). — Toby Speight, Oct 25 '19 at 12:27

mch · Answer 1 · 2019-10-25T07:51:48.383

4

while (p[i] != -1);

remove the trailing ;.

Also malloc gives you uninitialized memory, so chances are that the condition is not true at the first iteration. You probably want a do { ... } while(...); loop.

edited Oct 25 '19 at 07:51

answered Oct 25 '19 at 07:49

mch

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Oh yeah, thanks. I was first doing it with "do", then changed my mind, didn't notice that. – Chinguun Erdenebadrakh Oct 25 '19 at 07:51
But now, it's a Heap Corruption error. How do I fix this? – Chinguun Erdenebadrakh Oct 25 '19 at 07:52
You increase `n` and run a loop with the new value and access `p[i]`, which is `malloc`ed with the old value. – mch Oct 25 '19 at 07:54
`if (i <= 0)` and `printf("%d", p[i]);` looks dangerous. – mch Oct 25 '19 at 07:57

score 3 · Answer 2 · answered Oct 25 '19 at 07:49

Reformatting your code reveals the (or at least a) bug:

int i = 0;
int *p, *a;
int n = 1;
p = (int *)malloc(n * sizeof(int));
printf("Enter integers here, and input -1 when done:\n");
while (p[i] != -1)
  ;
{
  scanf_s("%d", &p[i]);
// ...

There's a stray semicolon after the first while, making it an infinite loop of nothing.

What makes a program not follow the loop condition?

2 Answers2