password strenght determination

[jrok]

hey, i am writing a program that will determine the strengh of any password entered, based on a certain criteria.
i have written the code and run it, but the way i wrote it, i always get "verify1 = n, verify2= n, verify3 = n. I know it is because, for example if I enter "Manypoepl&56them" as password, the do while loop sets verify 1 to 'y' the first time, but then sets it to 'n' when it checks the second letter. I want to know if anyone has any idea on how to avoid this. The code is below;

#include <iostream>
#include <cstdlib>
#include <fstream>
using namespace std;


int main()
{
int count=0;
char password, verify1, verify2, verify3, verify4;


cout << "Welcome.\n" <<
"This program helps youu to determine the strength\n" <<
"of any password you may chose, based on all of the following criteria for password strength:\n" <<
"1.At least 12 total characters.\n" <<
"2.At least one digit.\n" <<
"3.At least one lowercase letter.\n" <<
"4.At least one uppercase letter.\n" <<
"5.At least one character that is neither a letter nor a number,\n" <<
"(for example, a punctuation character).\n";

cout << "Now enter your choice of password:\n";

do
{
cin.get(password);

if (isupper(password))
verify1 = 'y';
else
verify1 = 'n';

if (islower(password))
verify2 = 'y';
else
verify2 = 'n';

if (isdigit(password))
verify3 = 'y';
else
verify3 = 'n';

if (isalpha(password))
verify4 = 'y';
else
verify4 = 'n';

count ++;
}while (password != '\n');

if ((verify1 == 'y' && verify2 == 'y') && (verify3 == 'y' && verify4 == 'n') && (count >= 12))
{
{
cout << "This password looks pretty good.\n";
}
}
else
{
if (verify1 == 'n')
cout << "Your password does not have at least one uppercase letter.\n" << endl;

if (verify2 == 'n')
cout << "Your password does not have at least one lowercase letter.\n" << endl;

if (verify3 == 'n')
cout << "Your password does not have at least one digit.\n"<< endl;

if (verify4 == 'y')
cout << "Your password does not have at least character that is neither\n"
"a letter nor a number.\n" << endl;
if (count < 12)
cout << "Your password should be at least 12 characters long.\n"<< endl;
}


while ((verify1 == 'n' || verify2 == 'n') || (verify3 == 'n' || verify4 == 'y') || (count < 12))
{
cout << "Please enter another password for verification.\n";
do
{
cin.get(password);
}while (password != '\n');
}

char wait; cin >> wait;
return 0;

}

[Timaster]

Start them all out as 'n' by default, and flip them to 'y' as soon as you find at least one of each.

[jrok]

i tried that, but the same thing happens again. Do you have any particluar logic of operation for the code in mind?

[Timaster]

Okay... your code has a staggering amounts of errors.

Here's how I'd model the loop:

program beginning, with variable declarations, intro text, etc.
main loop begin
	set everything to default state
	input password
	begin loop through every character
		4 if statements, set the corresponding verify char to 'y' if each of the requirements is found
	end character loop
	output results
main loop end
return 0;

Hopefully that helps. Give it an earnest go, but if that didn't make it clear, I'll write some actual code for you.

[jrok]

Thanks Timaster, i changed the code to the following and it works pretty well, the only twitch is that, it cannot verify a character that is neither a letter nor number. It always seems to get one such character no matter what I put in;
Any help?


#include <iostream>
#include <cstdlib>
using namespace std;

void verification(bool& try_newpassword, char& verify1, char& verify2, char& verify3, char& verify4, char& character);
//Precondition; This function requires the user to enter any password he/she choses for verification.
//Postcondition; The function then determines the strenght of the password based on the following criteria;
//1.At least 12 total characters.
//2.At least one digit.
//3.At least one lowercase letter.
//4.At least one uppercase letter.
//5.At least one character that is neither a letter nor a number,
//(for example, a punctuation character).
//If password satisfies all criteria, the function tells the user so,
//otherwise, it tells the user what criterion is not met by the password.


int main()
{
int count=0;
char password, verify1 = 'n', verify2 = 'n', verify3 ='n', verify4 ='y', character ='n';
bool try_newpassword;

cout << "Welcome.\n" <<
"This program helps youu to determine the strength\n" <<
"of any password you may chose, based on all of the following criteria for password strength:\n" <<
"1.At least 12 total characters.\n" <<
"2.At least one digit.\n" <<
"3.At least one lowercase letter.\n" <<
"4.At least one uppercase letter.\n" <<
"5.At least one character that is neither a letter nor a number,\n" <<
"(for example, a punctuation character).\n";


cout << "Please enter your choice of password:\n";

verification(try_newpassword, verify1, verify2, verify3, verify4, character);

while (try_newpassword == true)
{
cout << "Please enter another password for verification.\n";

verification(try_newpassword, verify1, verify2, verify3, verify4, character);
}


char wait; cin >> wait;
return 0;

}


void verification(bool& try_newpassword, char& verify1, char& verify2, char& verify3, char& verify4, char& character)
{
char password;
int count;
do
{
cin.get(password);

if (isupper(password))
verify1 = 'y';

if (islower(password))
verify2 = 'y';

if (isdigit(password))
verify3 = 'y';

if (isalpha(password))
verify4 = 'y';
else
character = 'y';

count ++;
}while (password != '\n');

if ((verify1 == 'y' && verify2 == 'y') && (verify3 == 'y' && character == 'y') && (count >= 12))
{
{
cout << "This password looks pretty good.\n";
}
try_newpassword = false;
}
else
{
if (verify1 == 'n')
cout << "Your password does not have at least one uppercase letter.\n" << endl;

if (verify2 == 'n')
cout << "Your password does not have at least one lowercase letter.\n" << endl;

if (verify3 == 'n')
cout << "Your password does not have at least one digit.\n"<< endl;

if (character == 'n')
cout << "Your password does not have at least character that is neither\n"
"a letter nor a number.\n" << endl;
if (count < 12)
cout << "Your password should be at least 12 characters long.\n"<< endl;

try_newpassword = true;


}

return;
}

[Timaster]

What you'll want to do for that is use isalpha() and isdigit() on every character, and set the flag to 'y' if it is neither. Use the logical and (&&), or (||), and not (!) operators (you'll need two).
For example: !(isalpha() || !isdigit()) is true when isalpha() is false and isdigit() is true.
Again, let me know if I'm just not making any sense, and I'll write you some actual code :)

[jrok]

Thanks again Timaster, I tried to use the logical statements above , but the results are the same as before, everything works, except the checking of the non-letter, non-digit character. Sorry to bother you again, but I think I might have to take you up on your offer and ask for some actual code. :)
Thanks.

[firedraco]

You could do it two ways, with an or, checking whether it is not an alpha or a digit, or you could check if it isn't an alpha and isn't a digit.

[Timaster]

Yep. And since I promised...

if (!isalpha(password)&&!isdigit(password))
	verify4 = 'y';

or

if (!(isalpha(password)||isdigit(password)))
	verify4 = 'y';

To me, it makes sense in English, as firedraco wrote it, but you can also prove it by looking through it and making sure that the result is what we want.
I'll post a table in a few minutes... have to go somewhere at the moment.

[jrok]

thanks firedraco. Timaster, thanks again. I looked at ur code and found my source of errorr. Seems like I was using a logical || instead of &&, so I was getting a true, even when one logic was false. I had completely missed it. Thanks again. It works perfectly now.

[jrok]

thanks firedraco. Timaster, thanks again. I looked at ur code and found my source of errorr. Seems like I was using a logical || instead of &&, so I was getting a true, even when one logic was false. I had completely missed it. Thanks again. It works perfectly now.

[Timaster]

Oh good. I was about to post a table, but if you got it, cool :)