当没有浮点数据类型时，为什么此代码会出现浮点异常？

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:

1. “Floating point exception” in code that contains no floats
2. Floating Point Exception C++ Why and what is it?

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

Answer 1

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.