Enter password, hide by "*".

I write a little program, just want to test getch() function.
When you enter password, they dont display in monitor,
instead of "*". It 's running OK.
But some statements look very stupid,
please give some improvement!
thanks.

PHP Code:

---

#include <iostream.h>
#include <conio.h>
#include <stdlib.h>

void main()
{
    char m[3]={'t','o','y'};   //to save original password.
    char n[3]={'0','0','0'};  //initialize your entering password.
    cout<<"Please enter 3 bit password: "<<flush;  
    n[0]=getch();            //enter word ,but no display it              
    cout<<"*"<<flush;    //when enter word, display “*”
    n[1]=getch();
    cout<<"*"<<flush;
    n[2]=getch();
    cout<<"*"<<flush;
    if(m[0]==n[0]&&m[1]==n[1]&&m[2]==n[2])   //stupid code!
    {
        cout<<"password correct,welcome!"<<endl; 
    
     }
    else
    {
        cout<<"error."<<endl;
    }

system("pause");    //pause running window.
return;

} 


---

it looks ok to me besides the

Code:

---

void main()

---

main should always return a int

Code:

---

int main(void)
{
  //code
  return 0;
}

---

@ prog-bman :
OH, it's a tip "int instead of void" or rule ?

How about this statement:

Quote:

---

if(m[0]==n[0]&&m[1]==n[1]&&m[2]==n[2]) //stupid code!

---

Thanks.

That looks fine.

>>OH, it's a tip "int instead of void" or rule ?
It is a rule. The C++ standard requires main to return an integer except in very specialized implementations where it imposes no restrictions. The general rule of thumb is that if you do not know whether or not you are using such an implementation, you are not and thus the relaxed rules do not apply to you. As such, instead of saying that every time somebody uses void as the return type for main, most helpful souls will simply tell you that returning void from main is illegal or wrong or evil or something similar for the sake of brevity.

@ Princeton :
Thanks,
I think I have learning a lot of valuable knowledge after entered this great forum.
Thanks every friend !

You will probably learn about control loops in the near future. They will provide you with the techniques to "clean up" the sections of the code that you felt uncomfortable with.

One other thing.
<iostream.h> is no longer standard, use <iostream> and the std namespace (putting 'using namespace std;' after your includes is the easiest to start with).
<stdlib.h> is standard C, but it is good to get in the habbit of using <cstdlib> in C++ along with the namespace.
Just be aware that <conio.h> is a non-standard header and you may or may not find it if you change compilers.

Code:

---

#include <iostream>
#include <conio.h>
#include <cstdlib>

using namespace std;

---

>>if(m[0]==n[0]&&m[1]==n[1]&&m[2]==n[2]) //stupid code!
An easier way to do this would be:

Code:

---

char m[4]="toy";  //4th character is '\0' (null).
char n[4]="000";  //same as above.

(...)

if(strcmp(m, n) == 0)  //compare the strings m and n
{...}
else
{...}

**OR**

char m[3] = {'t','o','y'};
char n[3] = {'0','0','0'};

(...)

if(memcmp(m, n, 3) == 0)  //compare 3 elements of m and n
{...}
else
{...}

---

Hope this helps! :)

ok your code corrected will be like this:

#include <iostream>
#include <cstdlib>
#include<conio.h>

using namespace std;

int main()
{

char m[3]={'t','o','y'}; //to save original password.

char n[3]={'0','0','0'}; //initialize your entering password.
cout<<"\n"
cout<<"\nPlease enter 3 bit password: "<<flush;

n[0]=getch(); //enter word ,but no display it

cout<<"*"<<flush; //when enter word, display “*”

n[1]=getch();

cout<<"*"<<flush;
n[2]=getch();
cout<<"*"<<flush;
if(m[0]==n[0]&&m[1]==n[1]&&m[2]==n[2]) //stupid code!
{
cout<<"password correct,welcome!"<<endl;

}
else
{
cout<<"error."<<endl;
}

system("pause"); //pause running window.
return 0;

}


but I ran it,mmm the pasword is toy!!!!

just to let you know that isn't a 3 bit password

I like the original code better. Slightly less standard, but it's code-tagged (Actually PHP tagged, but that's better than nothing), and the spacing is more intuitive. Abyssphobia, use code tags!

yeah it's 4 bit pasword!!! zero is null duh!!mmm sorry a little distraction
mmm code tags?
How?? I read the post but I didnt got it. =(
you know, Im gonna print the post to look it better=)

regards,

Abyss

No its a 3 bit. 0 - 1 - 10 - 11. 11==3.

what, no.

Code:

---

#include <limits.h>
number_of_bits=CHAR_BIT*(strlen(m));
number_of_bits+=CHAR_BIT;

---

plust your code doesn't have enough room for a terminating '\0'(if I enter three characters)

Just for your info, there's an example of how to obtain a password from the user here.

Sorry guys...looks like 24 bits to me. Maybe 3 bytes. Could be more.

Quote:

---

Originally Posted by C++Child

No its a 3 bit. 0 - 1 - 10 - 11. 11==3.

---

This looks compiler-specific to me. ;)

Quote:

---

yeah it's 4 bit pasword!!! zero is null duh!!mmm sorry a little distraction
mmm code tags?
How?? I read the post but I didnt got it. =(

---

Actually, it's 3 bytes (prize of a ++rep goes to pianorain!).

To use code tags, put [ code ] before your code and [ /code ] after your code (without the spaces, I'm just putting the spaces so that it doesn't think I'm actually using code tags ;)). Read the link in Salem's signature for more info.

Well, statistically speaking, it'll most likely be less than the equivalent of a 3-byte (24-bit) password. You are likely to have only 62 possible entries in those three spaces (a-z, A-Z, and 0-9), assuming you let the user type it in, instead of the maximum possible of 256 (2^8). So, instead of there being 256^3 (2^24) possible keys, there are 62^3. So, doing a bit of rounding so I don't need to pull out a calculator (62 ~ 64), we see that there are in fact ~64^3=2^18 possibilities, so what you have there is, in effect, an 18 bit (or 2.25 byte) password.

Sorry, I couldn't help myself. ;)

Well, technically you CAN input direct ASCII values for the password. Hold ALT and press 1,2,3 and then release ALT, and you'll end up with the character corresponding to ASCII 123. I've done that for several of my DOS programs before I realized that you could get the same effect from outputting (char)123 :) Although, on Windows there seems to be some oddity where alt-023 would give a different result from alt-23... Of course, if the user sets his/her own password, it should be no problem anyways.

Sorry, I couldn't help myself. ;)^2

ok now Im testing the code tags (thank you toysoldier

Code:

---


#include <iostream>
#include <cstdlib>
#include<conio.h>

using namespace std;

int main()
{

char m[3]={'t','o','y'}; //to save original password.

char n[3]={'0','0','0'}; //initialize your entering password.
cout<<"\n"
cout<<"\nPlease enter 3 bit password: "<<flush;

n[0]=getch(); //enter word ,but no display it

cout<<"*"<<flush; //when enter word, display “*”

n[1]=getch();

cout<<"*"<<flush;
n[2]=getch();
cout<<"*"<<flush;
if(m[0]==n[0]&&m[1]==n[1]&&m[2]==n[2]) //stupid code!
{
cout<<"password correct,welcome!"<<endl;

}
else
{
cout<<"error."<<endl;
}

system("pause"); //pause running window.
return 0;

}

---

I hope this works

Gee, how did you know my real name is toysoldier? :rolleyes:

**EDIT**
And congrats on your first successful code tag :)

>>Read the link in Salem's signature for more info.
I guess you mean mine?! :confused:

Sorry. I swear, it was a typo! *edits post in a lame attempt to cover up* :p

**EDIT**
Foiled. The edit button's gone on my post :rolleyes:

Code:

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if(m[0]==n[0]&&m[1]==n[1]&&m[2]==n[2]) //stupid code!

---

There's an easier way to compare strings in cstring. Use strcmp(). strcmp() returns a value of false (0) if both strings are the same. so say if(!strcmp(m[],n[])

>>so say if(!strcmp(m[],n[])

Code:

---

if (strcmp(m, n) == 0)

---

Or the equivalent but arguably less clear

Code:

---

if (!strcmp(m, n))

---

However, in this case because the arrays are not terminated by a null character they are not technically strings of any kind. As such, strcmp will cause undefined behavior because it will test beyond the boundaries of the array.

Even if the arrays were valid C-style strings you would still be wise to consider the overhead of a library function call over three simple equality tests. In some cases the difference may be enough to argue strongly in favor of the inline code.

mmm ok thx everybody, also Hunter thx,and toysoldier, I owe u one, thx 4 the patience
sorry If anyone get mad, Im sorry
well practicing the the tips you gave,
I just want to put how it looks 4 the newbies like me, Ive learned a lot!!!

Code:

---


                #include <iostream>
                #include <cstdlib>    // original #include<stdlib.h>
                #include <conio.h> 
 
                using namespace std;                       
                int main( )
           
              {

                      char m [4] = "toy";     
                      char n  [4] ="000";                       
                      cout<<" \n enter 3 bit password:"<<flush;
                 
                n [0] = getch( );  // getch( );
                      cout<<" * "<<flush; 
                     
                n[1]  = getch( );
                      cout<< "  * "<<flush;
                     
                  n[2]  = getch( );
                        cout<< "  * " <<flush;
             
        if (strcmp( m , n ) = = 0 ) 
                  {
                                  cout<<" password correcta, bienvenido"<<endl;
                  }
              else
              {
                              cout<<"error "<<endl;
              }
          system ("pause");   
        return 0;
      }

---

and

Code:

---


#include <iostream>
                #include <cstdlib> 
#include <conio.h> 
 
                using namespace std;                     
               
int main( )
           
              {

                      char m [4] = "toy";     
                      char n  [4] ="000";     
                 
                      cout<<" \n enter password:"<<flush;
                 
                n [0] = getch( ); 
                      cout<<" * "<<flush; 
                     
                n[1]  = getch( );
                      cout<< "  * "<<flush;
                     
                  n[2]  = getch( );
                        cout<< "  * " <<flush;
             
        if (strcmp( m , n ) = = 0 ) 
                  {
                                  cout<<" password correcta, bienvenido"<<endl;
                  }
              else
              {
                              cout<<"error "<<endl;
              }
          system ("pause");   
        return 0;
      }

---

ok guys tell me , what is the diference between the endl,flush like appears below

Code:

---

cout<<"whatever"<<endl;
cout<<"whatever"<<flush;
cout<<"whatever";

---

thx in adv.
love abyss

this is the princeton's way!!!

Code:

---


#include <iostream>
#include <cstdlib> 
#include <conio.h> 
 
                using namespace std;                     
               
int main( )
           
              {

                      char m [4] = "toy";     
                      char n  [4] ="000";     
                 
                      cout<<" \n enter password:"<<flush;
                 
                n [0] = getch( ); 
                      cout<<" * "<<flush; 
                     
                n[1]  = getch( );
                      cout<< "  * "<<flush;
                     
                  n[2]  = getch( );
                        cout<< "  * " <<flush;
             
        if (!strcmp( m , n )  ) 
                  {
                                  cout<<"this is the  password , welcome"<<endl;
                  }
              else
              {
                              cout<<"error "<<endl;
              }
          system ("pause");   
        return 0;
      }

---

>>what is the diference between the endl,flush
endl will print a newline and flush the stream, flush will not print a newline and flush the stream.

>>this is the princeton's way!!!
Well, it is not the way I would do it, but that will suffice. :D I would do something more like this if asked to use conio.h.

Code:

---

#include <cstdlib>
#include <fstream>
#include <iostream>
#include <string>
#include <vector>
#include <conio.h>

namespace {
  const std::string default_file("input.txt");
}

std::string decrypt(const std::string& pass);
void get_pass(std::string& buffer);
bool valid_pass(const std::string& buffer, const std::vector<std::string>& passwords);

int main(int argc, char *argv[])
{
  std::string buffer;
  std::vector<std::string> passwords;
  std::string pass_file(default_file.c_str());

  // Acquire filename and open file
  if (argc > 1) {
    pass_file.assign(argv[1]);
  }
  std::ifstream in(pass_file.c_str());
  if (!in.is_open()) {
    std::cerr << "Error opening password file" << std::endl;
    return EXIT_FAILURE;
  }

  // Load passwords into memory
  while (std::getline(in, buffer)) {
    passwords.push_back(decrypt(buffer));
  }
  buffer.clear();

  // Get user password and validate
  get_pass(buffer);
  if (valid_pass(buffer, passwords)) {
    std::cout << "Valid password" << std::endl;
  }
  else {
    std::cerr << "Invalid password" << std::endl;
  }

  return EXIT_SUCCESS;
}

std::string decrypt(const std::string& pass)
{
  // Placeholder for a password encryption scheme
  return pass;
}

void get_pass(std::string& buffer)
{
  int ch;

  while ((ch = getch()) != '\r') {
    if (ch == '\b') {
      // Handle backspacing intelligently
      buffer.erase(buffer.end() - 1);
      cprintf("\b \b");
    }
    else {
      buffer.push_back(static_cast<char>(ch));
      putch('*');
    }
  }
  putch('\n');
}

bool valid_pass(const std::string& buffer, const std::vector<std::string>& passwords)
{
  std::vector<std::string>::const_iterator it = passwords.begin();
  const std::vector<std::string>::const_iterator end = passwords.end();

  for (; it != end; ++it) {
    if (buffer == *it) {
      return true;
    }
  }

  return false;
}

---