My EVEN ODD Program giving odd output for every 10 digit even number

I created odd even program to detect odd and even numbers. It is working fine for all numbers less than 10 digits but when I give input of 2222222222 in the code, it gives the output of odd but how?

#include<iostream>
using namespace std;

int main(){

    int a;
    cin>>a;

    if(a%2==0){
        cout<<"This is EVEN";
    }
    else{
        cout<<"This is ODD";
    }

    return 0;
}

I tried giving input of 2222222222 and expected the output of even but it gave odd.

I guess you are using 32 Bit and thus, the integer ranges from 2 147 483 647 to -2 147 483 648. Use another datatype (e.g. long long) or switch to 64 Bit.

EDIT: As @wohlstad said in his comment, switching to 64 Bit might not solve the problem. Using another datatype is the way to go.

The problem is that 2222222222 is greater than the max value for int (i.e. 2147483647). Thus it causes an overflow and it results in a wrong output.

There are a few ways to solve this problem. The easiest one is to use long long which has a large range (up to 9223372036854775807 in both 32 and 64 bit X86 variants).

Like this:

#include <iostream>
#include <limits>

int main()
{
    long long int a;
    std::cout << "Enter a number in range "
              << 0
              << " to "
              << std::numeric_limits<decltype( a )>::max()
              << ": ";
    std::cin >> a;

    if ( a % 2 == 0 )
        std::cout << "This is EVEN\n";
    else
        std::cout << "This is ODD\n";
}

Sample output:

Enter a number in range 0 to 9223372036854775807: 2222222222
This is EVEN

Another way of doing it is to get the input as a std::string and then check to see whether or not its least significant digit is even.

This is an example:

#include <iostream>
#include <locale>
#include <string>
#include <string_view>

bool is_number( const std::string_view str )
{
    const auto loc { std::cout.getloc() };

    for ( const char c : str )
        if ( !std::isdigit( c, loc ) ) return false;

    return true;
}

int main()
{
    std::string a;

    do
    {
        std::cout << "Enter a number as big as you want: ";
        std::cin >> a;
    }
    while ( !is_number( a ) );

    if ( a.back() % 2 == 0 )
        std::cout << "This is EVEN\n";
    else
        std::cout << "This is ODD\n";
}

Sample output:

Enter a number as big as you want: 2222222222222222222222222222222222222222222222222222222222222222222222222222222222229
This is ODD

Now note that in the above solution, the condition if ( a.back() % 2 == 0 ) may seem weird because a.back() returns a char and then we're applying the modulo (%) operator on it. This is valid since the ASCII values for even decimal digits are even as well (e.g. '0' == 48, '2' == 50, '4' == 52). So in the end, the result is guaranteed to be correct.