Pseudocode problem

[NEOS]

This is my code and I just need to know if that's how I make it ask for the password 5 times, using the for loop.

% Write a pseudocode for a program that checks to see if a password is valid.
% The criteria for the password are listed below. The user must enter in the
% password and a message must be given if the password is valid or invalid.
% If it is invalid the program must allow the user to enter in a new one;
% and again, up to 5 times.

Turing:

/*  Variables password : String  For i : 1..5  Get password  Function Capletter (Cap:Int)  Check the ASCII Code Of password(1),password(2),password(3),password(4),password(5),password(6),password(7),password(8),password(9),password(10),password(11),password(12)  If The ASCII Of each Letter Equals Between 65 And 90 Then  Display Password is Valid Message  Else  Display Password is Invalid Message  The Result is If The password has at least one ASCII code  End Capletter  Function Oneletter (Lett:Int)  If the ASCII code For password (1) Equals Between 65 And 90, 97 And 122 Then  Display Password is Valid Message  Else  Display Password is Invalid Message  The Result is If The password has at least one ASCII code  End Oneletter  Function Onepun (pun:Int)  If the ASCII code For password (1) Equals Between 33, 63, 46 Then  Display Password is Valid Message  Else  Display Password is Invalid Message  The Result is If The password has at least one ASCII code  End Onepun  Function Onenum (num:Int)  If the ASCII code For password (1) Equals Between 48 And 57 Then  Display Password is Valid Message  Else  Display Password is Invalid Message  The Result is If The password has at least one ASCII code  End Onenum  If the Password Does not Have a Puncuation mark, One letter, one Number, a Capital letter, At Least 6 charecters And A max of 12 letters then  Display Password is Invalid Message  elsif Password Does Have a Puncuation mark, One letter, one Number, a Capital letter, At Least 6 charecters And A max of 12 letters then  Display Password Is Valid Message  end if    End for /*

That's my code and I just need to know if that's how I make it ask for the password 5 times, using the for loop.

[syntax_error]

var count.

[Insectoid]

You don't need to check the ascii values, you can do

code:

if pass[1] > A and pass[1]<Z then     pass [1] valid end if

[NEOS]

but if I don't check the acsii code then how will the computer know what's greater than A and less than Z?

[Tony]

because otherwise you are applying the same function to both sides of the equation

code:

pass[1] >= 'A'

is the same as

code:

ord(pass[1]) >= ord('A')

Though ultimately you are still checking if a character is in a certain numeric range, even if you don't explicitly specify the conversion.

[Insectoid]

Yeah, it does the same thing. It just reads better and requires less typing.

[DanielG]

Turing basically does the ascii conversion for you when comparing them.

[andrew.]

It would be something like this:

Constant:
password - The real password (string)

Variables:
passattempt - The password the user enters (string)
counter - A counter to check how many times you entered in a password (integer)

Main loop:
Clear screen
Print "Enter the password"
Get passattempt
Check if passattempt equals the password
If it does, then exit the loop
If it doesn't then add one to counter and say how may login attempts you used (counter out of 5)
Check if counter equals 5
If it does, then quit (not exit, terminate the program)

After the loop:
Whatever you want