The value of iterator not updating inside an if statement in a for loop

Question

I want to know why the value of i is not updating to i=0 when i == n-1, as I have written in the code. The bool flag is just there to check if the if statement was rendering at all or not (It is).

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        int n, ans = 0;
        scanf("%d", &n);
        int s[n];
        bool flag = false;
        for (int i = 0; i < n; i++)
        {
            scanf("%d", &s[i]);
        }
        for (int i = 0; i < n; i++)
        {
            s[i] -= 1;
            if (s[i] < 0)
                break;
            ans++;
            if (i == (n - 1))
            {
                i=0;
                flag = true;
            }
        }
        printf("\n%d\n", ans);
        printf("%d", flag);
    }

    return 0;
}

[How to debug small programs](https://ericlippert.com/2014/03/05/how-to-debug-small-programs/) — Eljay, Aug 28 '20 at 18:24
See also: [Why should I not #include ?](https://stackoverflow.com/q/31816095/3422102) Then add `-Wshadow` as a compiler option. — David C. Rankin, Aug 28 '20 at 18:30
`printf("%d", flag);` -- If this is truly C++, stop using `printf` and instead use `std::cout`. A `bool` is a C++ type, and `printf` has no idea what a `bool` is. Possibly using `%d` as a format specifier yields undefined behavior. — PaulMcKenzie, Aug 28 '20 at 18:32
@PaulMcKenzie There is neither undefined behavior using the format specifier %d. — Vlad from Moscow, Aug 28 '20 at 18:36
[Why should I not #include ?](https://stackoverflow.com/q/31816095/5910058) — Jesper Juhl, Aug 28 '20 at 18:36
[Why is "using namespace std;" considered bad practice?](https://stackoverflow.com/q/1452721/5910058) — Jesper Juhl, Aug 28 '20 at 18:37
Hello! Welcome! Can you specify the inputs you test this with, and the expected result. — g_bor, Aug 28 '20 at 20:02
Did you try commenting out that line to see what changed? While you probably don't want to draw firm conclusions based on such a test, it is probably a good starting point. — JaMiT, Aug 28 '20 at 20:59

score 0 · Answer 1 · answered Aug 28 '20 at 19:04

How do you know it's not doing what you want. I ran it and got this:

$ ./foo 1 2 3 4

5 1 $

The final value (1) is true. So it looks like it's doing it to me.

However, this is pretty terrible code. It's a really bad practice to muck with your iterator variable inside your loop like that. That's GOING to bite you. Furthermore, you're not going to go back to 0 -- you're going to go back to 1, because the next thing it does is increment the variable with your i++. So your if-statement sets i to 0 and your for-loop turns it into 1.

But this is just weird. If you need to run the loop multiple times, then wrap it in a larger loop.

Others have commented on printf. Personally, I disagree with them. There are times printf is completely valid. HOWEVER -- the way you're using it for such simple, unformatted stuff, isn't really one of them.

I changed the last printf to this:

cout << "Flag (raw): " << flag << ". Which means: " << (flag ? "True" : "False") << endl;

Output is this:

Flag (raw): 1. Which means: True

As you can see -- it's 1, which is not 0, so it's true.

The value of iterator not updating inside an if statement in a for loop

1 Answers1