How is this for loop terminated?

Question

Found as a user submission on leetcode for the problem Plus One.

vector<int> plusOne(vector<int>& digits) {
  for (int i=digits.size(); i--; digits[i] = 0)
      if (digits[i]++ < 9)
          return digits;
  digits[0]++;
  digits.push_back(0);
  return digits;
}

Normally there's something in the conditional like i >= 0; How is this for loop terminating and not accesing some -i element?

I should mention I really like the look and flow of the code, might even use something similar in a coding interview. I obviously would need to be able to speak on it though if I plan on using it! — Podo, May 31 '19 at 01:43
So anytime I'm using a for loop to decrement i, putting `i > =0` is just sugar and not necessary? — Podo, May 31 '19 at 01:48
***is just sugar and not necessary?*** For readability I would always use i >= 0 — drescherjm, May 31 '19 at 01:49
@drescherjm or [`i --> 0`](https://stackoverflow.com/questions/1642028/what-is-the-operator-in-c) — M.M, May 31 '19 at 01:55
for (size_t i=digits.size(); --i==0; digits[i] = 0) would be clearer. — mfnx, May 31 '19 at 09:01

score 2 · Answer 1 · edited May 31 '19 at 03:37

2

A for loop terminates when the second expression is false or when contextually converted to a bool value produces false. In C++, 0 is contextually converted to false; all other integers convert to true in a bool context. (Thanks to M.M.for the link.)

So at the beginning of each loop the expression i-- is evaluated. It's just an expression, so it will produce a result. If the result it produces is zero, the loop will end. If you haven't seen this expression before, here is how it works:

It both decrements i and returns the original value of i before it was decremented. So if i were 5, evaluating i-- will have the "result" 5, but as a side effect, i will be 4 after the evaluation.

The big picture: i is decremented by 1 each time, so assuming it starts off positive, it will get down to 0, at which time evaluating i-- will produce 0.

edited May 31 '19 at 03:37

Remy Lebeau

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answered May 31 '19 at 01:45

Ray Toal

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So a conditional must be included if there is some increment (like `i < 10`; i++`) but when you decrement the conditional is automatically greater than 0? – Podo May 31 '19 at 01:46
Any a value other than 0 is true – drescherjm May 31 '19 at 01:47
No conditional is needed, interestingly enough. `i--` always returns a value. If `i` were 5, say, then evaluating `i--` will produce the value 5, and _as a side effect_ cause `i` to be 4. – Ray Toal May 31 '19 at 01:47
Nice link, I will add that. – Ray Toal May 31 '19 at 01:48
Didn't know that was U.B., thanks! Will remove that and look into it, cool. – Ray Toal May 31 '19 at 01:52
@M.M. if I am reading [this page](https://en.cppreference.com/w/cpp/language/eval_order) correctly, then starting with C++ 11 you _do_ get guarantees that the value computation of a postfix increment is sequenced before its side-effect. Prior to C++ 11 we had the "sequence points" (remember those?) in which case there may be U.B. – Ray Toal May 31 '19 at 02:04

How is this for loop terminated?

1 Answers1