char(string) to integer - c++

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;

Answer 2

not sure if i understand your problem, but why not just change
char* m[3];
to
int m[3];
you'd probably want to redo the loop
int m[3];
for(i=0; i<3; i++)
{
cout<<i+1<<". Fak nomer: "<<endl;
cin>>m[i];
while (m[i] < 1000000 || m[i] >9999999)
{
cout<<"number is less/more than 7 digits. Try again:"<<endl;
cin>>m[i];
}
cout<<"As integer: "<< m[i];
}

Answer 3

It is difficult to understand exactly what you're trying to achieve.
If I understand it correctly, you're trying to get 3 7-character input strings (of integers) from the user, and then check each integer of the first string in turn against the same position character in the other 2 strings?
#include <iostream>
#include <string>
int main()
{
std::string m[3];
for(int i = 0; i < 3; ++i)
{
std::cout << i + 1 << ". Fak nomer: "<< std::endl;
do
{
std::cin >> m[i];
}
while (m[i].size() != 7);
}
// we have enforced each string to be 7 characters long, check each character in turn
for(int i = 0; i < 7; ++i)
{
// get the i'th character of each string, and subtract the ascii '0' character to convert to an integer
int a1 = m[0][i] - '0';
int a2 = m[1][i] - '0';
int a3 = m[2][i] - '0';
std::cout << "checking " << a1 << " against " << a2 << " and " << a3 << std::endl;
if (a1 % a2 == 0)
std::cout << a1 << " % " << a2 << " == 0" << std::endl;
if (a1 % a3 != 0)
std::cout << a1 << " % " << a3 << " != 0" << std::endl;
}
return 0;
}