Why does my code crash with a floating point exception?

Question

For reasons I can't understand, this crashes with a floating point exception.

This is surprising, because I do not appear to have any floating point operations in my code and I don't set i to 0. I've even added i * i != 0 to my code to make sure that this is the case.

Please could someone help me?

This is my code:

#include <iostream>

int main() {
    const int A = 42;
    int i;
    i * i != 0;
    for (i = 1; i < 99999999; i++) {
        if (A % (i * i) == 0) {
            std::cout << i << std::endl;
        }
    }   
}

[plead read this](http://stackoverflow.com/questions/1452721/why-is-using-namespace-std-considered-bad-practice) — Ed Heal, Mar 12 '17 at 21:42
line 2: `i * i != 0;` is useless. furthermore, i don't see you dividing in here. — Shark, Mar 12 '17 at 21:42
@NickA `i` should be initialized to 1 in the line above, but due to UB involved... thanks btw. — Shark, Mar 12 '17 at 21:48
@Nick A "modulo 0 raises a div by zero error doesnt it?" - no, it invokes "undefined behaviour" which means *anything* could happen. — Jesper Juhl, Mar 12 '17 at 22:09
You code crashes when i = 65536, so you're dividing by 4294967296, which is 2^32. Coincidence? I THINK NOT! It's overflowing, and ending up as 0. I'd add an answer, but it's closed so I can't ... — OMGtechy, Mar 12 '17 at 22:28
i added i*i !=0 thinking it might avoid 0 in the calculation — Luka Rusadze LukaRuso, Mar 12 '17 at 22:48
I'll edit the question to express that, although it doesn't do what you think it does — OMGtechy, Mar 12 '17 at 22:50

Iharob Al Asimi · Answer 1 · 2017-03-12T22:23:52.370

3

Your code is invoking undefined behavior, because

i * i != 0

appears before initializing i, thus you can understand your program's behavior because there is undefined behavior happening, and one of the things that could happen is that i == 0 at some point.

Also, the statement does absolutely nothing so it's pointless.

edited Mar 12 '17 at 22:23

answered Mar 12 '17 at 21:44

Iharob Al Asimi

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@JesperJuhl Yes, that's why one of the possible things is that `i == 0` at some point. – Iharob Al Asimi Mar 12 '17 at 22:15
`i * i != 0` - This is defined - it does nothing except doing the calculation and comparison - then throwing away the result – Ed Heal Mar 12 '17 at 22:19
Obviously. I'm just trying to say that that is not what your answer says. Terminoligy and proper language is important and your answer is not clear. That's all. – Jesper Juhl Mar 12 '17 at 22:19
@JesperJuhl Thank you for pointing it our, I appreciate your opinion. Is it clearer now? – Iharob Al Asimi Mar 12 '17 at 22:20
@Iharob Al Asimi The part I mostly object to is "behavior is explicable" - what does that mean? That's not proper english as far as I know. Do you perhaps mean "behaviour is undefined"? ?? – Jesper Juhl Mar 12 '17 at 22:22
this is not the problem, although the points are valid, see my other comments on the question :) – OMGtechy Mar 12 '17 at 22:31
i tried adding that so "i" doesn't somehow become 0. because somehow i was getting 0 in "i" from loop – Luka Rusadze LukaRuso Mar 12 '17 at 22:43

Why does my code crash with a floating point exception?

1 Answers1