获取错误浮点异常：8

Question

I have no idea why g++ doesn't like my code. It ran fine in java. Any insights would be greatly appreciated.

#include<iostream>

using namespace std;

bool isPrime(long number);

int main(){
const long number = 600851475143;
long max = 0;
for(long i= 0; i*i <= number; i++)
    if(number % i == 0 && isPrime(i))
        max = i;
cout<< max << endl;

return 0;
}

bool isPrime(long number){
if(number <= 1) return false;
if(number == 2) return true;
if(number % 2 == 0) return false;

for(long     i= 3; i*i <= number; i+=2)
    if(number % i == 0)
        return false;
return true;
}

Answer 1

const long number = 600851475143;

There is overflow, long can't hold that big number.

see this link

LONG_MAX is 2147483647

try:

const unsigned long long number = 600851475143;
unsigned long longmax = 0;

Edit:

You can't % against 0, i starts from 0

for(long i= 0; i*i <= number; i++)
           ^^
{
    if(number % i == 0 && isPrime(i))
               ^^^
{
   max = i;
   cout<< max << endl;
}

}

Minor change to a working version:

bool isPrime(unsigned long long  number);

int main(){

    const unsigned long long number = 600851475143;
    unsigned long long max = 0;
    for(long i = 1; i*i <= number; i++)
    {
        if(number % i == 0 && isPrime(i))
        {
            max = i;
            cout<< max << endl;
        }
    }
    return 0;
}

bool isPrime(unsigned long long  number)
{
    if(number <= 1) return false;
    if(number == 2) return true;
    if(number % 2 == 0) return false;

    for(unsigned long long i= 3; i*i <= number; i+=2)
    {
        if(number % i == 0)
        {
            return false;
        }
    }
    return true;
}

Answer 2

I don't see a floating point anywhere, but if I had to guess it's because it's due to overflow. Use unsigned long long or long long instead of regular long .

sizeof(long) on some compilers has evaluated to 4, similar to sizeof(int) , which means that the limit of long is 2147483647. long long is required by the C++ standard to be at least 64-bits, double that of long and int , which has a signed maximum of 9223372036854775807.

The error stems from your code: You're doing modulus by zero, which is wrong.

Consider doing this instead:

#include <iostream>

using namespace std;

bool isPrime(unsigned long long number);

int main(){
    const unsigned long long number = 600851475143;
    unsigned long long max = 0;
    for(unsigned long long i= 1; i*i <= number; i++)
        if(number % i == 0 && isPrime(i))
            max = i;
    cout<< max << endl;

    return 0;
}

bool isPrime(unsigned long long number) {
    if(number <= 1) return false;
    if(number == 2) return true;
    if(number % 2 == 0) return false;

    for(unsigned long long i= 3; i*i <= number; i+=2)
        if(number % i == 0)
            return false;
    return true;
}

Notice how i = 0 was changed to i = 1

Answer 3

For me, this bug came up when checking for integer overflow when doing a product:

#define INT_MIN -2147483648 // -2^31
#define INT_MAX  2147483647 //  2^31-1

int out=-1, x=-5;

if ((out > 0 && (x > INT_MAX/out || x < INT_MIN/out)) ||
    (out < 0 && (x < INT_MAX/out || x > INT_MIN/out))) {
  // what to do for overflow
} else {
  out *= x;
}

The issue is that because abs(INT_MIN) > abs(INT_MAX) , exactly when out=-1 , the condition INT_MIN/out was causing an overflow of int (because of 0 , there's not enough room to fit INT_MAX+1 in int ). I added a new condition to fix the floating point issue: out == -1 && (x > INT_MAX || x <= INT_MIN)