Question

I was wondering how to compare one array into two arrays but with some rules:
In my code I need to type 8 digits to make the cycle – OK, I did it. Now I need to compare the array = m[3] to two different arrays:

first array must be with (if m[i]%2==0) ...

second array must be with (if m[i]%2!=0) ...

So if I type from keyboard those three rows in my Main array (m[3]):

12345678
12345689
12344331

After typing them, I need to set the in those two different arrays and here I think I need to make the char(string) to integer to make the check with %, or to somehow do the check only on the last digit (it will work the same way).
So here goes the next step after typing the 3 rows:

arrA=12345678
arrB=12345689 12344331


#include <iostream>
#include <conio.h>
#include <string>
#include <cstdlib>
#include <stdlib.h>
using namespace std;
int main()
{
    int i,n;
    char m[3];
    for(i=1; i<=3; i++)
    {
        cout<<i<<". Fak nomer: "<<endl;
        do
        {
            cin>>m[i];
            gets(m);
        }
        while (strlen(m)!=7);
        cout<<"As integer: "<<atoi(m);
    }
}

Answer 1

From what I understand, you are trying to read three positive integers into an array called m, but you want to ensure the following:

• all three number have eight digits
• the first number (m[0]) is even
• the second number (m[1]) is odd

It would be a lot easier if m can be an array of integers, then a conversion from a string to an int is not required. To read the three integers from the console, use this:

// m is now an array of integers
int m[3];
// loops from m[0] to m[2]
for(int i = 0; i < 3; i++)
{
    cout<<i<<": "<<endl;
    do
    {
        cin>>m[i];
    }
    while (!(m[i] >= 1e7 && m[i] < 1e8));
    // m[i] must be greater than or equal to 10000000 and less than 100000000
    //  before continuing
}

By making m an array of integers, checking for even or odd becomes easier:

if (m[0] % 2 == 0)
    cout << "m[0] is even" << endl;
if (m[1] % 2 != 0)
    cout << "m[1] is odd" << endl;