How do these two conditions differ?

Question

I was wondering if you could clarify me the difference between the conditions while(k) and while(k > 0) when do they differ?

#include <iostream>
int main() {
    int k;
    std::cin >> k; 
    while(k) {
        std::cout << "Hello" << "\n";
        k--;
    }
}

#include <iostream>
int main() {
    int k;
    std::cin >> k; 
    while(k > 0) {
        std::cout << "Hello" << "\n";
        k--;
    }
}

Does this answer your question? [Casting int to bool in C/C++](https://stackoverflow.com/questions/31551888/casting-int-to-bool-in-c-c) — ph3rin, Feb 08 '21 at 20:09
The first version doesn't take negative `k` into account. Try entering `-1` and see what happens. (Note that signed overflow is undefined in C++). — rustyx, Feb 08 '21 at 20:11
@Fabiodlt In this context, you will get the same result, if `k > 0`. Infinite recursion in the first case if `k < 0` — Damien, Feb 08 '21 at 20:12
I mean, it does. It just matters on what you mean by "take into account." — sweenish, Feb 08 '21 at 20:12
@Fabiodlt check this link for understanding while loop: https://www.programiz.com/c-programming/c-do-while-loops — Nitai Chandra Banik, Feb 08 '21 at 20:12
As @Damien said, `while(k)` is equivalent to `while(k != 0)` which is different from `while(k > 0)` only if k can jump over zero and become negative - which can't happen in your example, however if you wrote a program to print k/2 lines of "Hello" and you did that with `k-=2;` instead of `k--` then there would be a difference when you entered an off number for k.. — Jerry Jeremiah, Feb 08 '21 at 20:14
@Fabiodlt Both versions run into [undefined behavior](https://en.cppreference.com/w/cpp/language/ub) if you enter something other than a number, `abc` for example. Additionally, the first version runs into UB if you enter a negative number such as `-1`. — dxiv, Feb 08 '21 at 20:17
@πάνταῥεῖ Not sure that's really a duplicate. Closely related, for sure ... but where is the cast in this question? — Adrian Mole, Feb 08 '21 at 20:27
@AdrianMole _"but where is the cast in this question?"_ It's done implicitely, and conditions require a `bool` value, isn't that the whole point? Also OP, confirmed, it explains their problem: https://stackoverflow.com/questions/66108612/how-do-these-two-conditions-differ?noredirect=1#comment116879720_66108612 (the auto generated duplicate comment was there before) — πάντα ῥεῖ, Feb 08 '21 at 20:30
@Adrian Well, as always, there's nothing wrong with duplicate duplicate questions, as long quesiton and answer(s) themselves provide valuable content. ;-) — πάντα ῥεῖ, Feb 08 '21 at 20:55

score 1 · Answer 1 · answered Feb 08 '21 at 20:18

1

The two code snippets you present will give equivalent results if – and only if – the input value for k is not negative. Try it.

Specifically, for the first snippet, if a value of -1 is input for k, the --k; line inside the while loop will (maybe) never reduce k to zero, or (maybe) just take a long time (until k reaches INT_MIN – something like -2147483648), depending on how the platform you are using handles signed integer underflow.

The while (k) loop will run until k is zero.

However, in the second snippet, a negative input for k will mean that the loop never runs (k will not be greater than zero on the first test).

answered Feb 08 '21 at 20:18

Adrian Mole

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I mean why in the first snippet the condition while(k) is regarded as while(k!=0)? – Fabiodlt Feb 08 '21 at 20:30
@Fabiodlt If you're happy that the linked duplicate answers your question, then fine. If I were to elaborate in my answer, I would simply quote that same question/answer. – Adrian Mole Feb 08 '21 at 20:38

How do these two conditions differ?

1 Answers1