Why does this code get floating point exception when there is no float data-type?

Question

I'm not dividing by zero and there is no float datatype in my code, I still get floating point exception.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    unsigned long long int t,n;

    cin>>t;
    while(t--)
    {
        cin>>n;
        unsigned long long int deno = pow(10,n-1),count=2,sum = 0,f1=1,f2=1;

         while(1){
            sum = f1+f2;
            f1 = f2;
            f2 = sum;
            count++;
            if((int)(sum/deno)>0){
                cout<<count<<endl;
                 break;
             } 
        }

    }
    return 0;
}

All the previous questions on the same had the similar problem of dividing by Zero but variable deno can never be zero as n>=2.

Previous research from my side:

Problem statement: https://www.hackerrank.com/contests/projecteuler/challenges/euler025/problem

It passes 2 test cases and fails 2. All are hidden test cases. Result image

On passing the input 1 50 we can reproduce the error. Details:

 GDB trace: Reading symbols from solution...done. [New LWP 15127] Core
 was generated by `solution'. Program terminated with signal SIGFPE,
 Arithmetic exception.
 #0  main () at solution.cc:23 
 23 if((int)(sum/deno)>0){
 #0  main () at solution.cc:23

floating point execption is when you attempt a division or modulo by 0. it isn't exclusive to floating points — Tyker, Jul 11 '18 at 22:15
How long does it take to reproduce the problem? When I run your code it just keeps running. — merlin2011, Jul 11 '18 at 22:15
Use your debugger. Watch the value of `deno` and see if it ever is zero. — alter_igel, Jul 11 '18 at 22:15
@merlin2011 It works perfectly on hackerrank.com Passed Test cases took less than a sec — Breakpoint, Jul 11 '18 at 22:16
@Easton Bornemeier hidden test case! This works fine for 2 test cases but produces floating point exception for rest two — Breakpoint, Jul 11 '18 at 22:19
@TotallyNoob, In this case, it would be helpful to provide a link to the original problem. — merlin2011, Jul 11 '18 at 22:20
https://www.hackerrank.com/contests/projecteuler/challenges/euler025/problem — Breakpoint, Jul 11 '18 at 22:22
I was unable to edit the question this is the problem statement — Breakpoint, Jul 11 '18 at 22:22
Your code seem exactly runnable without exception. Did you debug your code step by step? — Yılmaz edis, Jul 11 '18 at 22:25
@Yılmazedis debugging doesn't easily help if you don't know the input that leads tot he problem. — bolov, Jul 11 '18 at 22:26
@scohe001 Even after C++98 `std::pow()` only works with floating point numbers. — eesiraed, Jul 11 '18 at 22:27
@Yılmazedis Yes I tried to print values of deno and as much as I can think of deno never became zero hence floating point exception is a bit confusing in this case. — Breakpoint, Jul 11 '18 at 22:27
@FeiXiang Ahh looking closer at (7) it seems you're right. I skimmed and didn't look too closely at what `Arithmetic1` was thinking it'd handle `int` — scohe001, Jul 11 '18 at 22:29
@bolov thanks! I have edited the question and put the error message for better understanding — Breakpoint, Jul 11 '18 at 22:37
Do you realize that an `unsigned long long` can only go up about 19 digits while the problem requires 5000? Even a `long double` can only go up to 308 digits (with floating point error, of course). You're going to need a new approach. — eesiraed, Jul 11 '18 at 22:40
@FeiXiang Thank you so much! I'm going to change the approach. — Breakpoint, Jul 11 '18 at 22:44

AnT stands with Russia · Accepted Answer · 2018-07-11T23:57:21.697

9

It is perfectly normal for integer division to produce an exception that is reported as "floating point exception" on some platforms (Linux, for one example). You can easily get it from integer division by zero, or, for another example, by triggering overflow as in

int i = INT_MIN;
int b = -1;
i = i / b;

http://coliru.stacked-crooked.com/a/07c5fdf47278b696

In certain contexts this exception might appear or disappear depending on optimization levels. The exception is normally only triggered when the compiler decided to generate the actual division instruction (as opposed to optimizing out the division).

In your case unsigned integer division is used, so division by zero seems to be the only possible culprit. I would guess that this

unsigned long long int deno = pow(10,n-1);

happens to result in zero in deno. pow is a floating-point function that produces a floating-point result. Conversion from floating-point type to integer type leads to undefined behavior if the original value is too large (which is the case for n equal to 50). Note that this is the case even if the target integer type is unsigned.

edited Jul 11 '18 at 23:57

answered Jul 11 '18 at 22:31

AnT stands with Russia

312,472
42
525
765

1

Thank you, but how should I change my code to overcome this problem? – Breakpoint Jul 11 '18 at 22:42
Thank you for looking deep into the problem and helping me out – Breakpoint Jul 11 '18 at 22:54
yes, just checked for `1 50` `deno` is `0`. On Visual Studio however is `9223372036854775808`. This is indeed UB. – bolov Jul 11 '18 at 22:55
1

Write your own pow() function. `std::pow` always converts ints and returns floating point results. There is no equivalent for ints only in the standard library. There are numerous int equivalents on the web. – doug Jul 12 '18 at 05:44
with n>=19 the logic is ill-formed on a wide range of platforms. it just overflows the 64 bit boundries of long long. – Red.Wave Jul 12 '18 at 09:37
Mildly related: You can use the "Russian peasant algorithm":http://lafstern.org/matt/col3.pdf for efficient integer exponentiation. It is based on an ancient approach of _strength reduction_ of multiplication to addition/subtraction/comparison. In this case, it reduces operation strength from exponentiation to at most multiplication. – Arne Vogel Jul 16 '18 at 09:56

Why does this code get floating point exception when there is no float data-type?

1 Answers1