cout numbers with all digits pair numbers

[catalin10]

Write an algorithm that displays the numbers that have all their digits pair. The numbers will be introduced by the user, the program will end when 0 is introduced.

#include <iostream>
using namespace std;
main ()
{
    int num, test, digit, impair_digit;
    impair_digit=0;
    while(num!=0)
    {
        cin>>num;
        test=num;
        while(test!=0)
        {
            digit=test%10; //extract each digit of the number and see if its pair
            test/=10;
            if(digit/2!=0)//if digit isn't pair add 1 to the counter
            impair_digit++;
            else
            impair_digit=impair_digit;
            }

            if(impair_digit==0)//if the counter is not 0 then at least 1 digit is impair
            cout<<"\n the number "<<num<<" has only pair digits";
            else
            cout<<"\n the number "<<num<<" doesn't have only pair digits";
    }
} 

When I run the program, no matter what number I introduce, it always returns that the number introduced doesn't have only pair digits, even if i enter 2222.

[vin]

if(digit/2!=0) does not test whether a digit is even.

For example 2/2 is 1 which is not 0 so your program considers 2 not an even number.

[Moschops]

if(digit/2!=0)//if digit isn't pair add 1 to the counter

To reiterate vin, this is functionally identical to:

if (digit != 0 && digit!= 1 %% digit != -1) // integer division 1/2 is zero

[catalin10]

Silly me was being careless; made it work with if(digit%2!=0). Thanks.

[evanc9606]

Please mark as soved if it is solved