C++ Help, Please. Thank You.

[Happy Monkey]

I think I was wrong about the bad behavior again; it should work. But it is very complicated. The test functions can return false when the password has been changed to a good one. The structure of the while loop corrects for it, but the behavior of the functions is not what one would expect.

The snippet I posted would give the same behavior, but is much simpler.

This would allow the error reporting to be in the loop, and only calls each test once:

bool passLength(string) // just return true/false.  Don't read a new pw or print error
bool containDigit(string) // just return true/false.  Don't read a new pw or print error

string getPW()
{
  string password;
  bool ok=false;
  while ( ! ok )
  {
    cout << "Please enter a password: ";
    cin.getline(password, SIZE);

    if (! passlength(password)
    {
      cout << "Passwords must be at least 6 characters long" << endl;
    } 
    else if ( ! containDigit(password) )
    {
      cout << "Passwords must include at least on digit (1-9)" << endl;
    }
    else
    {
      ok=true;
    }
  }
  return password;
}

[Flint]

Originally Posted by HM

The test functions can return false when the password has been changed to a good one.

If it is a good password the function cannot return false.

Originally Posted by HM

The structure of the while loop corrects for it, but the behavior of the functions is not what one would expect.

No, man. The if condition corrects for it by not calling the function unless it is false. The while loop doesn't correct the behavior, the while loop just exits the loop before the next iteration.



In a nutshell, this is how I check for two conditions:

while !(a) or !(b)
{
if !(a)
call (a)

if !(b)
call (b)
}

This will repeat until the conditions are met. It will also exit immediately when the conditions ARE met.

[tw]

Originally Posted by Flint

while !(a) or !(b)
{
if !(a)
call (a)

if !(b)
call (b)
}

Only reading the summary code: Are (a) and (b) a function or method? Or are each a property?

If (a) or (b) are functions: when a bad user name is entered in
while ( !(a) ....
That causes the while loop to drop to an 'If' condition. Then another
(a) function is executed.

Then the while loop again again executes (a) again:
while ( !(a) or !(b) )
Everytime (a) and (b) are executed - an new entry occurs. IOW what happened in "call (a)" gets replaced by another execution of "while ( !(a) ..."


To work, the code should read something like this:


do
{call (a)
if ((a).result == good)
{call (b)}
}
loop until ( (a).result == good and (b).result == good )


(a).result and (b).result are properties.
call(a) and call(b) are methods.

Summary code many not accurately reflect what your actual code does. But the difference between an executed function call(a) and its resulting property (a).result should be clearer. As I read the original summary code, everytime (a) is executed, a previous user name (good or bad) is erased and a new username is entered.

[Flint]

@tw: (a) and (b) are bool functions. while !(a) does not cause if !(a) to drop. while !(a) OR !(b) simply makes the decision to enter the loop or not.

Originally Posted by tw

Everytime (a) and (b) are executed - an new entry occurs.

Correct. That is what makes this work. The loop continuously evaluates, "falling through" true conditions and "catching" on false conditions. false conditions always prompt for a new password and immediately reevaluate the same condition. The loop continues to evaluate until it is able to "fall through" completely (exit the loop).

Here is the program running:

Please enter a password: a
Passwords must be at least 6 characters long
Please enter a password: abc
Passwords must be at least 6 characters long
Please enter a password: abcdef
Passwords must include at least on digit (1-9)
Please enter a password: 123
Passwords must be at least 6 characters long
Please enter a password: abcdef
Passwords must include at least on digit (1-9)
Please enter a password: a1
Passwords must be at least 6 characters long
Please enter a password: abc123
Thank you that is a valid password

Press Enter to close window...

Here is the code:

#include<iostream>
#include<cstring>
#include<cctype>
using namespace std;
	
	//Function checks the password for length.		(a)
	bool passLength(char[]);

	//Function checks the password for a digit.		(b)
	bool containDigit(char[]);
	
	const int SIZE = 21;
	char password[SIZE];

int main()
{
	cout << "Please enter a password: ";
	cin.getline(password, SIZE);
	
	while ((!passLength(password)) || (!containDigit(password)))  
		{
	if	(!passLength(password))
		(passLength(password));						//(a)

	if	(!containDigit(password))
		(containDigit(password));					//(b)	
		}

	cout << "Thank you that is a valid password" << endl;
	
	
	//Keep the window open until Enter key is pressed.
	cout << "\nPress Enter to close window..." << endl;
	std::cin.get();

	return 0;
}	

	bool passLength(char password[])				//(a)
{
	int lengthPass = 6;
	int length = strlen(password);
	
		if (lengthPass <= length)
			return true;
		else 
			{
			cout << "Passwords must be at least 6 characters long" << endl;
			cout << "Please enter a password: ";
			cin.getline(password, SIZE);
			return false;
			}
}


	bool containDigit(char password[])				//(b)
{
	int index = 0;
	int length = strlen(password);
	
	for (index = 0; index < length; index++ )
	{
		if (isdigit(password[index]))
			return true;
	}
		cout << "Passwords must include at least on digit (1-9)" << endl;
		cout << "Please enter a password: ";
		cin.getline(password, SIZE);		
		return false;
	
}

[Happy Monkey]

Originally Posted by Flint

If it is a good password the function cannot return false.

If the password going in is bad, abd the password entered in the function is good, then the function returns false, even though the password is good.

No, man. The if condition corrects for it by not calling the function unless it is false.

It calls it no matter what. If that first call returns false, it calls the function a second time (third if you count the loop test).

Every time you call if (!a()), you are calling a().

The while loop doesn't correct the behavior, the while loop just exits the loop before the next iteration.

While giving the user two more chances to change the password.

[Flint]

Originally Posted by HM

Every time you call if (!a()), you are calling a().

Perhaps, but this is not the behavior I see while using Visual C++ 2010 Express.

Please enter a password: abc
Passwords must be at least 6 characters long
Please enter a password: abc123
Thank you that is a valid password

Press Enter to close window...

Another example:

Please enter a password: abc
Passwords must be at least 6 characters long
Please enter a password: abcdef
Passwords must include at least on digit (1-9)
Please enter a password: 123
Passwords must be at least 6 characters long
Please enter a password: abcdef
Passwords must include at least on digit (1-9)
Please enter a password: abc123
Thank you that is a valid password

Press Enter to close window...

[Perry Winkle]

Originally Posted by Flint

Perhaps, but this is not the behavior I see while using Visual C++ 2010 Express.

Maybe you're getting confused output because you're not flushing the output stream (i.e., flush(cout). Another, but less likely, answer might be that the function call is getting optimized away.

Either way you should at least store the values of a and b each time through the loop.

[Flint]

Originally Posted by Happy Monkey

If the password going in is bad, abd the password entered in the function is good, then the function returns false, even though the password is good.

I'm interested to know why you think this is; as I cannot conceive of a logic where this makes sense. Aside from the fact that this isn't what happens. I'm just interested to hear your reasoning.

[Happy Monkey]

Originally Posted by Flint

I'm interested to know why you think this is; as I cannot conceive of a logic where this makes sense. Aside from the fact that this isn't what happens. I'm just interested to hear your reasoning.

Say you call containDigit("abc")

    int index = 0;
    int length = strlen(password);
    
    for (index = 0; index < length; index++ )
    {
        if (isdigit(password[index]))
            return true;
    }

It gets through these steps, and gets here:

        cout << "Passwords must include at least on digit (1-9)" << endl;
        cout << "Please enter a password: ";
        cin.getline(password, SIZE);

The user enters "abc1".
And then:

        return false;

The password is now "abc1", and containDigit() returned false.

You don't see the problem in your execution because your while loop corrects for the behavior of the test functions. But if you tested the functions individually, you would see it.

Originally Posted by Flint

This code only "calls" passLength IF passLength is false.

What exactly do you mean by "IF passLength is false"? A function can only be false if it is called and executed.

Unlike the solutions graciously suggested by Pete Zicato and Happy Monkey, in my code I do not require a superfluous bool variable. I use the bool function to evaluate itself. Maybe this is "wrong" but it works.

My first suggestion also doesn't need the bool, without the side effects in the test functions.

[Flint]

When I call containDigit("abc")...

Please enter a password: abcdef
Passwords must include at least on digit (1-9)
Please enter a password: abc
Passwords must be at least 6 characters long
Please enter a password: abc123
Thank you that is a valid password

Press Enter to close window...

If a condition is met, he loop "toggles" to the next false condition. If you DON'T meet the condition, you are stuck at that point until it is met (calling that same bool function with your input). Once you enter the correct input, the loop falls through to evaluate the next condition. Repeat. No matter what you input into my program, you can't make it break, and you can't make it behave wrongly. It works with 100% accuracy.

Again, the while function simply decides whether to enter the loop or not at each iteration.

A function can only be false if it is called and executed.

That seems to make sense, but running my program suggests otherwise.

I find it hard to argue with the fact that what I've done works. Because it works.

You don't see the problem in your execution because your while loop corrects for the behavior of the test functions. But if you tested the functions individually, you would see it.

I am interested to test this, but not right now, as I am about to leave to go to our cabin in the country where we have no internet and no cell phone service! Thanks for your thoughts and conversation on this--I am new to this and it really helps me to bounce ideas off of people such as yourself that I trust are logical. :::checking out:::

[Perry Winkle]

Originally Posted by Flint

I find it hard to argue with the fact that what I've done works. Because it works.

Your solution is perfectly acceptable for a classroom assignment. But you might get your ass kicked in a code review for settling for "it works."

In the real world, you want functions to have as few side-effects as possible. Ideally, a function would have no side-effect other than its return value. Some languages even ENFORCE this restriction.

Your solution depends on side effects. This makes it harder to reason about how and why it works. Happy Monkey's does not depend on side effects, and is easier to grok.

(I hope this post doesn't come off as critical. I'm just trying to add a little more to think about.)