Why is this not working?
case ARGV.length
when 0
abort "Error 1"
when > 2
abort "Error 2"
end
It's not valid ruby syntax.
What you need is
case
when ARGV.length == 0
abort "Error 1"
when ARGV.length > 2
abort "Error 2"
end
When you write case x, the important part you need to understand is that ruby takes the x and then applies a comparison to the argument or expressions you insert in the when clause.
The line where you say when x >2 reads to ruby like:
if ARGV.length == > 2
When you remove a specific object from the case statements, you can apply conditionals within the when statements .
Use 1.0 / 0.0 to get infinity, which fixes #mosch's code:
case ARGV.length
when 0
raise "Too few"
when 3..(1.0/0.0)
raise "Too many"
end
You don't have to be Chuck Norris to divide by a floating point zero.
Well, it doesn't work because it's not valid ruby syntax. However, you can do this:
x = 15
case x
when 0..9 then puts "good"
when 10..12 then puts "better"
when 13..200 then puts "best"
else
puts "either great or poor"
end
An if statement would probably be more fitting for your code, since you don't have a definitive range/value, but rather just a greater-than:
if ARGV.length == 0
abort "Error 1"
elsif ARGV.length > 2
abort "Error 2"
end
You can use either if elsif or lambda/proc (then notation is just for shortness):
case x
when 1 then '1'
when ->(a) { a > 2 } then '>2'
end
You can also create a named variable for clarity:
greater_than_2 = ->(a){ a > 2 }
case x
when 1 then '1'
when greater_than_2 then '>2'
end
You can also create your own comparison predicate:
greater_than = ->(n) { ->(a) {a > n} }
case x
when 1 then '1'
when greater_than[2] then '>2'
when greater_than[5] then '>5'
end
Answering your initial question when > 1 is not a valid syntax, since > is binary operator in Ruby, which is to say its arity 2, meaning it takes 2 arguments. Don't be confused of term binary vs bitwise.
Related
I wrote a simple guess the number game. But it keeps looping even when I input the correct number. Please help, thanks!
puts "Pick a number between 0 - 1000."
user_guess = gets.chomp.to_i
my_num = rand(831)
guess_count = 0
until user_guess == my_num do
if user_guess == my_num
guess_count += 1
puts "you got it!"
elsif user_guess <= 830
guess_count += 1
puts "higher"
else user_guess >= 1000
guess_count += 1
puts "lower"
end
end
puts "You guessed my number in #{guess_count} attempts. Not bad"
The part of the code that asks for a number from the user is outside the loop, so it will not repeat after the answer is checked. If you want to ask the user to guess again when their guess is wrong, that code needs to be inside the loop.
my_num = rand(831)
guess_count = 0
keep_going = true
while keep_going do
puts "Pick a number between 0 - 1000."
user_guess = gets.chomp.to_i
if user_guess == my_num
guess_count += 1
puts "you got it!"
keep_going = false
elsif user_guess <= 830
guess_count += 1
puts "higher"
else user_guess >= 1000
guess_count += 1
puts "lower"
end
end
puts "You guessed my number in #{guess_count} attempts. Not bad"
This code still has some bugs in it that stops the game from working correctly though, see if you can spot what they are.
As #Tobias has answered your question I would like to take some time to suggest how you might make your code more Ruby-like.
Firstly, while you could use a while or until loop, I suggest you rely mainly on the method Kernel#loop for most loops you will write. This simply causes looping to continue within loop's block until the keyword break is encountered1. It is much like while true or until false (commonly used in some languages) but I think it reads better. More importantly, the use of loop protects computations within its block from prying eyes. (See the section Other considerations below for an example of this point.)
You can also exit loop's block by executing return or exit, but normally you will use break.
My second main suggestion is that for this type of problem you use a case statement rather than an if/elsif/else/end construct. Let's first do that using ranges.
Use a case statement with ranges
my_num = rand(831)
guess_count = 0
loop do
print "Pick a number between 0 and 830: "
guess_count += 1
case gets.chomp.to_i
when my_num
puts "you got it!"
break
when 0..my_num-1
puts "higher"
else
puts "lower"
end
end
There are a few things to note here.
I used print rather than puts so the user will enter their response on on the same line as the prompt.
guess_count is incremented regardless of the user's response so that can be done before the case statement is executed.
there is no need to assign the user's response (gets.chomp.to_i) to a variable.
case statements compare values with the appropriate case equality method ===.
With regard to the last point, here we are comparing an integer (gets.chomp.to_i) with another integer (my_num) and with a range (0..my_num-1). In the first instance, Integer#=== is used, which is equivalent to Integer#==. For ranges the method Range#=== is used.
Suppose, for example, that my_num = 100 and gets.chomp.to_i #=> 50 The case statement then reads as follows.
case 50
when 100
puts "you got it!"
break
when 0..99
puts "higher"
else
puts "lower"
end
Here we find that 100 == 50 #=> false and (0..99) === 50 #=> true, so puts "higher" is displayed. (0..99) === 50 returns true because the integer (on the right of ===) is covered by the range (on the left). That is not the same as 50 === (0..90), which loosely reads, "(0..99) is a member of 50", so false is returned.
Here are a couple more examples of how case statements can be used to advantage because of their reliance on the triple equality method.
case obj
when Integer
obj + 10
when String
obj.upcase
when Array
obj.reverse
...
end
case str
when /\A#/
puts "A comment"
when /\blaunch missiles\b/
big_red_button.push
...
end
Use a case statement with the spaceship operator <=>
The spaceship operator is used by Ruby's Array#sort and Enumerable#sort methods, but has other uses, as in case statements. Here we can use Integer#<=> to compare two integers.
my_num = rand(831)
guess_count = 0
loop do
print "Pick a number between 0 and 830: "
case gets.chomp.to_i <=> my_num
when 0
puts "you got it!"
break
when -1
puts "higher"
else # 1
puts "lower"
end
end
In other applications the spaceship operator might be used to compare strings (String#<=>), arrays (Array#<=>), Date objects (Date#<=>) and so on.
Use a hash
Hashes can often be used as an alternative to case statements. Here we could write the following.
response = { -1=>"higher", 0=>"you got it!", 1=>"lower" }
my_num = rand(831)
guess_count = 0
loop do
print "Pick a number between 0 and 830: "
guess = gets.chomp.to_i
puts response[guess <=> my_num]
break if guess == my_num
end
Here we need the value of gets.chomp.to_i twice, so I've saved it to a variable.
Other considerations
Suppose we write the following:
i = 0
while i < 5
i += 1
j = i
end
j #=> 5
j following the loop is found to equal 5.
If we instead use loop:
i = 0
loop do
i += 1
j = i
break if i == 5
end
j #=> NameError (undefined local variable or method 'j')
Although while and loop both have access to i, but loop confines the values of local variables created in its block to the block. That's because blocks create a new scope, which is good coding practice. while and until do not use blocks. We generally don't want code following the loop to have access to local variables created within the loop, which is one reason for favouring loop over while and until.
Lastly, the keyword break can also be used with an argument whose value is returned by loop. For example:
def m
i = 0
loop do
i += 1
break 5*i if i == 10
end
end
m #=> 50
or
i = 0
n = loop do
i += 1
break 5*i if i == 10
end
n #=> 50
1. If you examine the doc for Kernel#loop you will see that executing break from within loop's block is equivalent to raising a StopIteration exception.
I have this loop:
puts "Welcome to the Loop Practice Problems"
puts " Write a number between 1 and 10, but not 5 or else...."
ans = gets.chomp!
if ans < 1
puts "Tf bruh bruh"
elsif ans > 10
puts "Now you just playin"
elsif x == 5
print "You wildin B"
else
puts "Fosho that's all I require"
end
It doesn't run properly, and I'm trying to understand why. If you can help me with this, I'd appreciate it.
If you know a good site for practice problems, I'd love to try it. I checked out Coderbyte and Code Kata, but the way they're set up doesn't look right, and they don't have questions to solve for fundamentals.
The issue here is that you're not converting ans to a number, but you're comparing it to one. ans is going to be a string.
In Ruby, when you compare a number to a string, Ruby says that the two aren't equal:
"1" == 1
=> false
You can reproduce the problem with this code:
puts "Welcome to the Loop Practice Problems"
puts " Write a number between 1 and 10, but not 5 or else...."
ans=gets.chomp!
p ans
The p method will output an "inspected" version of that object, (it's the same as doing puts ans.inspect). This will show it wrapped in quotes, which indicates that it's a string.
You probably want to do this:
ans = gets.chomp!.to_i
The to_i method here will convert the number to an integer, and then your comparisons will work correctly.
You have to convert input string type object into integer type
ans = gets.chomp!.to_i #input string convert into integer.
if ans < 1
puts "Tf bruh bruh"
elsif ans > 10
puts "Now you just playin"
elsif x == 5
print "You wildin B"
else
puts "Fosho that's all I require"
end
def Summer
#summer = true
puts "Your fruit are ripe for the picking."
if #tree_age == 1..5 && #tree_age > 0
#oranges = 5
elsif #tree_age == 6..15
#oranges = 20
else
#oranges = 50
end
end
I'm trying to ensure a tree between a certain age range gives x oranges, however I'm stuck with the following error referring to my elsif statement:
Orange_tree.rb:14: warning: integer literal in conditional range
I have also tried using an if greater than && less than conditional statement, can somebody please explain what this error means, and how to reach my solution.
You have a few problems:
You'll want to put your ranges in parenthesis when other operators or methods are nearby. Your current error comes from Ruby parsing elsif #tree_age == 6..15 differently than you expect - it's treating it as (1 == 6)..15, and false..15 obviously doesn't make any sense.
To test a number is within a range, use (1..5) === num, not num == (1..5). Range#=== is defined to test that the Range includes the right hand side, while Fixnum#== and Fixnum#=== both just test that the right hand side is numerically equivalent.
You don't need to test #tree_age > 0. You're already testing that it's in 1..5.
You could also consider a case statement for this, which can be a bit easier to read. case does its comparisons using ===.
#oranges = case #tree_age
when 1..5 then 5
when 6..15 then 20
else 50
end
You should use include? instead of == to determine if the given number is within the range:
def Summer
#summer = true
puts "Your fruit are ripe for the picking."
if (1..5).include?(#tree_age) && #tree_age > 0
#oranges = 5
elsif (6..15).include? #tree_age
#oranges = 20
else
#oranges = 50
end
end
==:
Returns true only if obj is a Range, has equivalent begin and end
items (by comparing them with ==), and has the same exclude_end?
setting as the range.
Which is obviously not the case.
The problem is with the lines that say == with a range.
if ( 10 == 1..11) # throws integer literal in conditional range warning
puts "true"
end
If you did this instead
if ( 10.between?(1, 11))
puts "true"
end
I am starting to learn programming and I have chosen to learn Ruby using Codecademy. However, I was trying to consolidate my learning but I just can't get this to work!
print "What is 2 + 2 ="
sum_var = gets.chomp
sum_var.downcase!
if sum_var == "four" || 4
puts "Correct!"
else sum_var != "four" || 4
puts "Wrong! #{sum_var} is not the answer!"
end
It just returns 'Correct!' even if it is wrong.
You need write the code as below :
print "What is 2 + 2 ="
sum_var = gets.chomp
# don't need to apply the bang method like you did - sum_var.downcase!
if sum_var.downcase == "four" || sum_var == '4'
puts "Correct!"
else # else don't need condition checking, so I removed.
puts "Wrong! #{sum_var} is not the answer!"
end
sum_var = gets.chomp gives you a string, no where you are converting it to a number. So, evenif you are passing number from the console it became "4" or "7" etc.
Let me explain you also why always you got "Correct!" as a output
sum_var == "four" || 4 - In this expression, whenever sum_var is not equal to "four", your first expression was evaluated to false, but when control went to test the second expression, it found there 4. You know in Ruby all objects are true, except nil and false. So 4 is considered as true. Thus in your code always if block was getting executed, and you were keep getting as the output "Correct!".
Now in your code, some other mess you did, that I corrected in my above code.
The problem with your code lies in the line
if sum_var == "four" || 4
The == will usually return false, so the second part will be evaluated because the precedence of == is higher than the precedence of ||. Since all objects except false and nil are "truthy" in Ruby the expression will end up being true regardless of the users input. Correction as proposed by #ArupRakshit, just wanted to add some more reason to it.
This is a much simpler example of what I'm trying to do in my program but is a similar idea. In an, if statement how do I say not equal to?
Is != correct?
def test
vara = 1
varb = 2
if vara == 1 && varb != 3
puts "correct"
else
puts "false"
end
end
Yes. In Ruby the not equal to operator is:
!=
You can get a full list of ruby operators here:
https://www.tutorialspoint.com/ruby/ruby_operators.htm.