using isdigit to validate user input c++

[george08.08]

Hello. I am trying to validate a customer number using isdigit. If the user accidently enters a character other than an integer, I want it to allow the user to try again. The problem is it only will check the first number and if that is a digit it it presumes it is ok. I want it to check all the numbers entered. I have it currently looping around the whole thing so I can keep testing input. Thank you (in anticipation)

#include "stdafx.h"
#include <iostream>
#include <cctype>
#include <iomanip>

using namespace std;


int main()
{
char password [6]; //password is 6 digit long
int i;

cout << "enter your login : " ;
cin >> password;

for (i = 0; i < 6; i++)
{
	if (isdigit (password[i]))
	{cout << "\nThank you." << endl;}// it is an integer yay!
	else
		cout << "\nYou can only have a numeric login" << endl;
	cout <<"Enter your login: " << endl;
	cin >> password;
}
system("pause");
	return 0;
}

[Ancient Dragon]

First, if you want a 6-digit password then the variable must be declared as 7 characters to allow room for the null-terminator that cin will add.

There are several ways to do this (and more efficient ones too), here is one of them.

bool flag = false;
for(i = 0; i < 6; i++)
{
  if( !idigit(password[i]) )
  {
    cout << "Wrong password\n";
    flag = true;
    break;
  }
}
if( flag)
{
  // go back and ask for password again
}

[sundip]

#include "stdafx.h"
#include <iostream>
#include <cctype>
#include <iomanip>

using namespace std;

bool validate(char[] password, int len)
{
    for (i = 0; i < len; i++)
    {
	if (!isdigit (password[i]))
        return false;
    }
    return true;
}

int main()
{
    char password [7]; //password is 6+1('/0') digit long
    int i;

    cout << "enter your login : " ;
    cin >> password;

    while(!validate(password,7))
    {
        cout << "\nYou can only have a numeric login" << endl;
        cout <<"Enter your login: " << endl;
        cin >> password;
    }
    cout << "\nThank you." << endl;}// it is an integer yay!
    system("pause");
    return 0;
}

[george08.08]

Thank you, since I am very new to this, all variations are appreciated. I still have the problem where it will not pick up if I have entered a letter. for example, if I enter 1234k I will get an error message, I feel because it is not 6 characters long, but if I type 12345k It will allow it through because it is the right length even though it has a letter. I have written ...

while (password < MIN_NUM && password > MAX_NUM)
{cout << "Your password must be 6 digits long." << endl;
cout <<"Enter your 6 digit password.: " ;
cin >> password;

to ensure they only type in 6 digits, I just need each of those digits validated, presumably with a loop, but I don't know how to implement it. Thank you again.

[murnesty]

normally i do that with string. So user can enter whatever length/alpha/number they want. After that only write errorcheck function for the input.

[george08.08]

Is it also possible to do it with atoi (I think that was what it is)?

[george08.08]

Thank you all, your input has put me on the right track. I could not use len, since it also evaluated the null at the end which always returned and error. This is the finished code. Thanks again

#include "stdafx.h"
#include <iostream>
#include <cctype>
#include <string>
#include <iomanip>
 
using namespace std;
 
int main()
{
    char password [7]; //password is 6+1('/0') digit long
	int len;
    bool validate(char[],int);
	//int strlen(char password[]);

	cout << "enter your login : " ;
    cin >> password;

	len = strlen(password); //This will get the exact length of the password typed in
  
    while(!validate(password, len))
    {
        cout << "\nYou can only have a numeric login" << endl;
        cout <<"Enter your login: " << endl;
        cin >> password;

		len = strlen(password);
     }
    cout << "\nThank you." << endl;
	system("pause");

}
	

	bool validate(char pass[], int len)
{
	int i = 0;
	
    for (i = 0; i < len; i++)
    {
	if (!isdigit (pass[i]))
        return false;
    }
    return true;
}