[c++] hide char

[Dang_1]

hi everyone, i need some help.
This is my problem.
When I input a string to the program, I want the screen console print out '*'.

ex:
string s;
cout<<"password :";
getline(cin,s);

when i type "abc123" for the password. The screen console will show "******".

I'm sorry about my poor english.

[Suzie999]

I don't know how to do this, so this wont be a great help but might get you started.

I recall seeing a similar question asked regarding windows programming and as I recall the API "SetConsoleMode" was used in the answer.

[Ancient Dragon]

There is no standard C or C++ function/class that will do what you want. So you can resort to non-standard functions, for example the functions in conio.h if your compiler supports that header file and library.

Have youy searched google for password code? like this link?

Another example here

[rubberman]

It is simple enough to write your own function to do this. Pass in the password to the function, and have it generate a string of '*' characters of the same length that it returns. Call this function string obfuscatePswd(const string& pswd). Then your code becomes this: cout << obfuscatePswd(password) << " :";

[Ancient Dragon]

That's easy enough, but the problem is getting the string from the keyboard without actually displaying the plain text, display stars instead of the text.

[Suzie999]

Perhaps it's possible to implement some sort of low level keyboard hook to capture the input but print "*".

[Dang_1]

i think i will use _getch().
if _getch()!='\n' i will cout<<"*";
but i can't backspace if i type wrong pass

[Ancient Dragon]

Here's how to backspace

int main()
{
    int c;
    while ((c = _getch()) != '\n')
    {
        if (c == '\b')
            putchar(c);
        else
            putchar('*');
    }
    return 0;
}

[iamthwee]

This is not possible. It is OS dependant

[Ancient Dragon]

what is not possible??

[iamthwee]

Sorry I worded it wrong. I meant in c++, as what he is asking for is OS dependant. Clearly, depending on your OS you can do this.

[Dang_1]

i used this

char* s1 = new char[100];
char y = _getch();
s1[0] = y;
int i = 0;

while (s1[i] != 13)
    {
        cout << '*';
        i++;
        y = _getch();
        s1[i] = y;
    }
s1[i] = '\0';

Pass = s1;
delete []s1;

but it is not perfect

[Ancient Dragon]

Didn't you read this post to show how to backspace?

line 16: if Pass in line 15 is a pointer then you can't delete s1 on line 16 because it will invalidate Pass also.