Taking password as input in C++

There are two methods in which input can be taken in a more secure way:

• Do not display any content.
• Display a special character such as an asterisk instead of actual content.

In this method, input content will be invisible. This can be implemented in two ways:

using <windows.h>:

Program 1:

Below is the program where console mode is set to enable, echo input, and reset the console mode:

C++

// C++ program to take the input
// invisibly
#include <iostream>
#include <windows.h>
using namespace std;

// Function take password and
// reset to console mode
std::string takePasswdFromUser()
{
    HANDLE hStdInput
        = GetStdHandle(STD_INPUT_HANDLE);
    DWORD mode = 0;

    // Create a restore point Mode
    // is know 503
    GetConsoleMode(hStdInput, &mode);

    // Enable echo input
    // set to 499
    SetConsoleMode(
        hStdInput,
        mode & (~ENABLE_ECHO_INPUT));

    // Take input
    string ipt;
    getline(cin, ipt);

    // Otherwise next cout will print
    // into the same line
    cout << endl;

    // Restore the mode
    SetConsoleMode(hStdInput, mode);

    return ipt;
}

// Driver Code
int main()
{
    string input;
    cout << "@root>>> ";

    // Function Call
    input = takePasswdFromUser();

    // Print the input
    cout << input << endl;
}

Output:

using <conio.h>:

For this getch() is used. This function takes a character input from user without buffer and doesn’t wait for the user to press “return” key.

Program 2:

Below is the C++ program to demonstrate the use of getch() in conio.h:

C++

// C++ program to demonstrate the
// use of getch()
#include <conio.h>
#include <iostream>
using namespace std;

// Function using getch()
std::string takePasswdFromUser()
{
    string ipt = "";
    char ipt_ch;
    while (true) {
        ipt_ch = getch();

        // Check whether user enters
        // a special non-printable
        // character
        if (ipt_ch < 32) {
            cout << endl;
            return ipt;
        }
        ipt.push_back(ipt_ch);
    }
}

// Driver Code
int main()
{
    string input;
    cout << "@root>>> ";

    // Function call
    input = takePasswdFromUser();
    cout << input << endl;
}

Output:

Drawback: The user can’t clear the response made earlier. When backspace is pressed, the input is returned.

Program 3:

Below is the C++ program to demonstrate the solution to the above drawback: 

C++

// C++ program to demonstrate the
// solution of above drawback
#include <conio.h>
#include <iostream>
using namespace std;

// Enumerator
enum TT_Input {

    // ASCII code of backspace is 8
    BACKSPACE = 8,
    RETURN = 32
};

// Function accepting password
std::string takePasswdFromUser()
{
    string ipt = "";
    char ipt_ch;
    while (true) {
        ipt_ch = getch();

        if (ipt_ch < TT_Input::RETURN
            && ipt_ch != TT_Input::BACKSPACE) {
            cout << endl;
            return ipt;
        }

        // Check whether the user
        // pressed backspace
        if (ipt_ch == TT_Input::BACKSPACE) {

            // Check if ipt is empty or not
            if (ipt.length() == 0)
                continue;
            else {

                // Removes last character
                ipt.pop_back();

                continue;
            }
        }
        ipt.push_back(ipt_ch);
    }
}

// Driver Code
int main()
{
    string input;
    cout << "@root>>> ";

    // Function call
    input = takePasswdFromUser();
    cout << input << endl;
}

Hiding the password by a special character(*):

The idea is to use the library <conio.h> here to hide password with asterisk(*). Below is the C++ program using conio.h to hide the password using *:

Program 4:

C++

// C++ program to hide the password
// using *(asterik)
#include <conio.h>
#include <iostream>
using namespace std;

// Enumerator
enum IN {

    // 13 is ASCII for carriage
    // return
    IN_BACK = 8,
    IN_RET = 13

};

// Function that accepts the password
std::string takePasswdFromUser(
    char sp = '*')
{
    // Stores the password
    string passwd = "";
    char ch_ipt;

    // Until condition is true
    while (true) {

        ch_ipt = getch();

        // if the ch_ipt
        if (ch_ipt == IN::IN_RET) {
            cout << endl;
            return passwd;
        }
        else if (ch_ipt == IN::IN_BACK
                 && passwd.length() != 0) {
            passwd.pop_back();

            // Cout statement is very
            // important as it will erase
            // previously printed character
            cout << "\b \b";

            continue;
        }

        // Without using this, program
        // will crash as \b can't be
        // print in beginning of line
        else if (ch_ipt == IN::IN_BACK
                 && passwd.length() == 0) {
            continue;
        }

        passwd.push_back(ch_ipt);
        cout << sp;
    }
}

// Driver Code
int main()
{
    string input;
    cout << "@root>>> ";

    // Function call
    input = takePasswdFromUser();
    cout << input << endl;
}