struggling with my pseudocode - pseudocode

I'm about to build a program written in pseudocode. I've already done most of the work , but I'm stuck on the code and I don't know what to do exactly. im a begginer and not everything is clear to me ...in one of the tasks i have to do , i have to make the program ask for the players name , which will be stored as a string then the program has to check if it exceeds the limit between 2/20 characters and inform the user if the input is wrong . i have researched and tried to figure out how i might be able to fix my code but i have a really short amount of time left and coudn't find anything regarding my problem :/ . this is the code ive done for this specific task. i know its wrong but i just dont know how to fix it . any help with be much appreciated . Thanks in advance :)
pseudocode:
// Getting user's name
valid = false
loop until valid is equal to true
Output" please enter your name "
Input playName
If (playName is => 1)AND(=<20)then
Valid = true
Otherwise
output "name exceeds the character limit"

I'm not sure what the syntax of your pseudo code is but :
assuming tabulation has meaning, you may have forgot to indent some lines to include them in the loop
'valid' is first declared with a lower case first letter so you may continue referencing it same way in line "Valid = true" -> "valid = true"
In the 'If' you want to test the lenght of the String and not compare the string to an int so maybe call a function length(String) that would return the length of the string or access a string.length attribute (as you wish in pseudo code)
You want the playName to be superior or equal to 2 "length(playName) >= 2" and inferior or equal to 20 "length(playname) <= 20"
The commonly used keyword meaning Otherwise is 'Else' as in
IF (Condition) THEN (code) ELSE (code)
I may modify you code like this :
// Getting user's name
valid = false
loop until valid is equal to true
Output" please enter your name "
Input playName
If (length(playName) >= 2) AND (length(playName) <= 20)
Then
valid = true
Else
output "name exceeds the character limit"

Related

Why isn't Lua reevaluating io.read("*n")?

I have some code
::redo::
io.write("input: ")
var = io.read("*n")
if var then
if var > 5 and var < 10 then io.write("yes\n") goto redo
else io.write("invalid\n") goto redo end
else io.write("invalid\n") goto redo end
that is supposed to check a numeric input value and return if it's within a certain range. If it isn't a numeric value, it's supposed to "redo" the script and ask for input again. The issue is that whenever it takes an input that isn't a number it repeats io.write("input: ") and io.write("invalid\n") unceasingly meaning it's skipping the var = io.read("*n") line. Is there a special meaning or quirk to io.read("*n") that keeps it from reevaluating? The code seems to work if replaced with io.read()
When you call io.read('*n') and it doesn't find a number, it doesn't use up the input, and any calls to io.read('*n') will read the same input over and over. You need to eat up the input and discard it by calling io.read('*l'). That will let you read new input with io.read('*n').
Another method would be to read a line with io.read('*l'), extract a number from it with string.match and convert it to a number with tonumber. Then you don't have to read the same input twice, but you'd have to decide what types of number notation you want to match. (io.read('*n') accepts various types of numbers, including hexadecimal and scientific notation.)

Why do `while` loops never check user input, whereas `if` statements do?

An if statement such as:
if first_name || first_name.length == 0
puts "You can't leave the first name blank, try again: "
first_name = gets
end
functions correctly, because it asks for user input, and when tested by not entering any input (including spaces), it prints an error message, followed by prompting for user input, following which, However, when I attempt to use the same in a while loop:
while first_name || first_name.length == 0
puts "You can't leave the first name blank, try again: "
first_name = gets
end
the loop doesn't function correctly.
The loop does the following instead:
• Print the error message above (OK)
• Asks for user input (OK)
• As soon as user input is received, the loop executes once again (Wrong).
It isn't testing the length of the user's (now correct) input, and then allowing the rest of the application to carry out its tasks.
I also tried unless, nil?, and so on to rectify the loop's error using an if statement inside the loop, with no success. I thought the while loop would evaluate first_name again once the error was corrected, and find that its length was no longer zero, however, that turned out not to be the case.
In Java, to say something isn't equal to something else (in terms of strings, for example), I can use ! before the variable name. Can I do this in Ruby?
You can try this:
while first_name.length == 0 do
puts "You can't leave the first name blank, try again: "
first_name = gets
end
The way while cycle works is it evaluates the statement after while keyword. If it is true then it runs the block. And after that just the same. In your case statement is first_name || first_name.length == 0 which always will be true as first_name is always present. That's the reason it will loop again each time whatever you would input.
I think your condition is just wrong : the condition
first_name || first_name.length == 0
Will be true if first_name is non-nil, this will always be the case since you are setting it to a string.
After some further digging on the Internet, I gave .chomp a go, to see what it would do, as I found out that when gets is used, it takes your input, but apparently also inserts an \n escape sequence, when Enter is pressed, meaning that my loop didn't function as required.
By using gets.chomp, in conjunction with .empty? I was able to remove this escape sequence from my input, and have the loop function correctly.
Here's the correct code, which prevents an infinite loop:
first_name = gets.chomp
while first_name.empty? do
puts "You can't leave the first name blank, try again: "
first_name = gets.chomp
end

Does ruby's case statement fall through?

I am writing a hangman game in ruby and I wanted to use a case statement to determine which body part to place corresponding to a number of incorrect guesses. I made this game using a board class I use for other games like chess and connect-4 because I have a method which serializes the board class allowing me to save and load the game without any extra code. For the game to be saved, I needed some way of determining the number of incorrect guesses for the hangman without adding extra variables to the board class. To solve this I used an instance variable on the board class called history, which can be used to push moves from the game to the boards history. When the board gets serialized, the history is saved as well, which can be read by the game and used to determine incorrect guesses.
In the hangman game, I have a method called read history (which I use for all the games since it solves the serialization issue described above). The read_history method is responsible for reading the past guesses, display them, and determine the number of incorrect guesses. This number is then passed to a hang method which determines which body parts of the hangman to add.
def hang(incorrect)
case incorrect
when 0
#hangman = [" ", " ", " "]
break
when 7
#hangman[2][2] = '\\'
when 6
#hangman[2][0] = '/'
when 5
#hangman[2][1] = '*'
when 4
#hangman[1][2] = '\\'
when 3
#hangman[1][0] = '/'
when 2
#hangman[1][1] = '|'
when 1
#hangman[0][1] = 'o'
end
end
If I were writing this in java, and a value of 5 were passed to the above method, it would read the statement until it hit "when 5" or in java terms "case 5:". It would notice that there is not a break in the statement and will move down the list executing the code in "case 4:" and repeating until a break is found. If 0 were passed however it would execute the code, see the break, and would not execute and other statements.
I am wondering if Ruby is capable of using case statements the way java does in the way that they fall through to the next statement. For my particular problem I am aware that I can use a 0.upto(incorrect) loop and run the cases that way, but I would like to know the similarities and differences in the case statement used in ruby as opposed to the switch-case used in java
No, Ruby's case statement does not fall through like Java. Only one section is actually run (or the else). You can, however, list multiple values in a single match, e.g. like this site shows.
print "Enter your grade: "
grade = gets.chomp
case grade
when "A", "B"
puts 'You pretty smart!'
when "C", "D"
puts 'You pretty dumb!!'
else
puts "You can't even use a computer!"
end
It's functionally equivalent to a giant if-else. Code Academy's page on it recommends using commas to offer multiple options. But you can still won't be able to execute more than one branch of logic.
It does not fall through.
Ruby just doesn't have the same behavior as Java for this type of statement.
If you want to simulate the fall through behavior, you can do something like this:
def hang(incorrect)
#hangman = [" ", " ", " "]
#hangman[2][2] = '\\' if incorrect > 6
#hangman[2][0] = '/' if incorrect > 5
#hangman[2][1] = '*' if incorrect > 4
#hangman[1][2] = '\\' if incorrect > 3
#hangman[1][0] = '/' if incorrect > 2
#hangman[1][1] = '|' if incorrect > 1
#hangman[0][1] = 'o' if incorrect > 0
#hangman
end

Can't convert String into Integer (TypeError)

I am learning Ruby
I am trying to create a simple script that will convert a given number to roman numerals (old style roman numerals)
I am unable to understand why I get the "can't convert String into Integer (TypeError)"
def convert_to_roman number
romans_array = [[1000,'M'],[500,'D'],[100,'C'],[50,'L'],[10,'X'],[5,'V'][1,'I']]
converted_array = []
romans_array.each do |rom_num|
num = rom_num[0]
letter = rom_num[1]
if number > num
times = number / num
roman_letter = letter*times
converted_array.push(roman_letter)
number = number % num
end
end
converted_array.join()
end
number = ''
puts 'please write a number and I will convert it to old style Roman numerals :)'
puts 'p.s. to exit this program simply hit enter on an empty line, or type 0 and enter :)'
while number != 0
number = gets.chomp.to_i
puts convert_to_roman number
end
My code is at:
https://github.com/stefanonyn/ruby-excercises/blob/master/roman_numerals.rb
You will see that at the end of the file commented out there is an old revision of the code, which actually does work but has a lot of repetition.
I would appreciate if someone could clarify why I get the error described above.
Please don't write the code for me, I am trying to learn Ruby, I would appreciate just some support in moving to the next step.
Thank you very much!
You are missing a comma in your array
romans_array = [[1000,'M'],[500,'D'],[100,'C'],[50,'L'],[10,'X'],[5,'V'][1,'I']]
^ here
This error is definitely not all that helpful, but the reason that it is appearing is that to the interpreter it looks like you are attempting to access a range of indexes in the [5,'V'] array for the last element. However the index's that are being provided go from 1 to 'I' which of course makes no sense. If it had been written [5,'V'][1,1] the last element of the array would be ['V'], which might have been even more confusing to debug!

Ruby: String Comparison Issues

I'm currently learning Ruby, and am enjoying most everything except a small string comparason issue.
answer = gets()
if (answer == "M")
print("Please enter how many numbers you'd like to multiply: ")
elsif (answer. == "A")
print("Please enter how many numbers you'd like to sum: ")
else
print("Invalid answer.")
print("\n")
return 0
end
What I'm doing is I'm using gets() to test whether the user wants to multiply their input or add it (I've tested both functions; they work), which I later get with some more input functions and float translations (which also work).
What happens is that I enter A and I get "Invalid answer."The same happens with M.
What is happening here? (I've also used .eql? (sp), that returns bubcus as well)
gets returns the entire string entered, including the newline, so when they type "M" and press enter the string you get back is "M\n". To get rid of the trailing newline, use String#chomp, i.e replace your first line with answer = gets.chomp.
The issue is that Ruby is including the carriage return in the value.
Change your first line to:
answer = gets().strip
And your script will run as expected.
Also, you should use puts instead of two print statements as puts auto adds the newline character.
your answer is getting returned with a carriage return appended. So input "A" is never equal to "A", but "A(return)"
You can see this if you change your reject line to print("Invalid answer.[#{answer}]"). You could also change your comparison to if (answer.chomp == ..)
I've never used gets put I think if you hit enter your variable answer will probably contain the '\n' try calling .chomp to remove it.
Add a newline when you check your answer...
answer == "M\n"
answer == "A\n"
Or chomp your string first: answer = gets.chomp

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