macOS Continued... C++ password help!

pilotError said:

The forum does that...
The other way to do it would be to get the characters and either echo a character back or don't echo at all.

Code:

char password[20];  // Accept up to 20 char password 

for (i=0; i<20; i++)
{

   password[i] = getch();
   cout << "*";

   if (ch[i]==13)  // should this be a 10?  I forget my CR's from my LF's...
      checkPassword();
}

Click to expand...

#include <iostream.h>
#include <stdlib.h>
#include <time.h>
#include <stdio.h>
#include <string.h>
#include <windows.h>
#include <conio.h>

int PassCheck(char password2[]);
void main()
{
	char password[20]; 
    for (int i=0; i<20; i++)
	{
       password[i] = getch();
       cerr << "*";
       if(password[i] == 9) 
       PassCheck(password);
	}
}
int PassCheck(char password2[])
{
	int check;
	int numtries = 0;
	cerr << "PLEASE TYPE IN THE PASSWORD...\nPASSWORD: ";
	cin >> password2;
	cerr << "LOGGING IN";
	Sleep(1000);
	cerr << ".";
	Sleep(1000);
	cerr << ".";
	Sleep(1000);
	cerr << ".\n";
	Sleep(1000);
	strcmp (password2, "xxxx");
	check = 1;
	if (check == 1)
	{  
	   cerr << "PASSWORD VERIFIED...\n";
	   cout << "goto op here lol";
	}
	else
	{
	numtries = numtries + 1;
	if (numtries == 5)
      {
	  cerr << "SYSTEM LOGGING OUT...\n";
      return 0;
	  }
	}
	cerr << "PASSWORD DENIED, PLEASE TRY AGAIN...\n";
	return 0;;
}

ChrisBrightwell said:

Your logic flow is messed up.

Try this:
1. Print "enter username".
2. Begin loop,
2a. accept input.
2b. compare it to the output
3c. either go back to 2a or end the loop
4. ???
5. PROFIT!

Click to expand...

pilotError said:

Try something like this...
I didn't test it, but logically its a little closer to what you need
Obviously it doesn't test for using the delete key so type carefully

You should really separate the password code into a different method (ie. getPassword() ) and return a success or failure

Code:

#include <iostream.h>
#include <stdlib.h>
#include <time.h>
#include <stdio.h>
#include <string.h>
#include <windows.h>
#include <conio.h>

int PassCheck(char password2[]);
int doWork();


void main()
{
	char password[20]; 
	int retryCount=0, retVal;
	
	do {
		cout << "PLEASE TYPE IN THE PASSWORD...\nPASSWORD: ";
	
	    for (int i=0; i<20; i++)
		{
	       password[i] = getch();
	       cout << "*";
	       if (password[i] == '\n')
				break;
		}
		
		retVal = PassCheck(password);
		
		if (retVal == 1)
		{
			cout << "PASSWORD VERIFIED...\n";
		   	doWork();
		}
		else
		 if ((retryCount + 1) < 5)
			cout << "PASSWORD DENIED, PLEASE TRY AGAIN...\n";
		
    } while (++retryCount < 5);

	cout << "TOO MANY RETRIES, SYSTEM LOGGING OUT...\n";
	exit(1);
}


int PassCheck(char password2[])
{

	if ( strcmp (password2, "xxxxx") == 0)
		return(1);
		
	return(0);
}

int doWork()
{
	cout << "Dude Buy a Mac!" << endl;
}

Click to expand...

asciimov said:

Hi Unknown,

Your looking in the wrong place to do what you want.

When you type in something your c program is not echoing your key strokes to the console, the console is.

The solution to this is to turn off echoing when your typing in your password, then back on when you exit that part of the program.

I don't know what console/shell your using, so I cant give you the specific command, but this info should be able to help you google-fu your solution.

Hope that helps,
asciimov

Click to expand...

ChrisBrightwell said:

Your logic flow is messed up.

Try this:
1. Print "enter username".
2. Begin loop,
2a. accept input.
2b. compare it to the output
3c. either go back to 2a or end the loop
4. ???
5. PROFIT!

Click to expand...

for (int i=0; i<20; i++)
		{
	       password[i] = getch();
	       cout << "*";
	       if (password[i] == '\n')
				break;
		}