Have written some test code for a program, trying to pass 2 values, a file and a number. The below doesn't work at all, but if I have something like puts "test" (outside the case) it works.
def read_album(music_file, number)
puts number #(this does something)
case number.to_i
when(number > 1)
puts "done"
when(number < 2)
puts "something"
when(number == 3)
puts "none of this inside case is working"
end
end
def main()
a_file = File.new("albums.txt", "r")
print("Choose album id: ")
choice_of_album = gets().to_i
read_album(a_file, choice_of_album)
end
main()
Your cases are not doing what you think. The expressions given to when are evaluated and the result will be compared to the case value using the case equality operator ===. An expression such as number > 1 will evaluate to either true or false. It makes no sense to compare this result to an integer in Ruby.
You should compare against the constants directly.
case number
when 1
# handle 1
when 2
# handle 2
when 3
# handle 3
else
# handle unknown case; error?
end
Note that other classes may override === to provide useful behavior. The Range and Regexp classes, for example, do this.
case number
when 1..3
# handle 1, 2 and 3
end
case string
when /pattern/
# handle pattern
end
Notably, the Proc class also does this!
def greater_than(n)
proc { |x| x > n }
end
case number
when greater_than(2)
# handle number > 2
end
You need to drop the number.to_i from the case statement.
Or do something like
case number.to_i
when 1..2
puts "foo"
when 3..100
puts "bar"
else
puts "foobar"
end
end
From the Ruby docs
Case statements consist of an optional condition, which is in the position of an argument to case, and zero or more when clauses. The first when clause to match the condition (or to evaluate to Boolean truth, if the condition is null) “wins”, and its code stanza is executed. The value of the case statement is the value of the successful when clause, or nil if there is no such clause.
Your version would evaluate to somehting like
if (number > 1) === number.to_i
and since you are comparing a number with a boolean expression this will not evaluate to true. If you had an else in the case statement this would have been called.
Related
I'm quite new to Ruby. So I have a method that takes string input and then compares it several times. The problem is - no matter what - it always go to the first case. For example:
def method (s)
if s <=> "help"
help()
elsif s <=> "add"
puts"Enter 2 numbers"
a = STDIN.gets
b = STDIN.gets
add(a,b)
else
err("Invalid command, type 'help' for list of available commands")
end
end
class.method(STDIN.gets)
always do "help()", no matter what the input is. How it should be written?
<=> is useful for ordering, it'll give you 0 when strings are equal, -1 when s is before "help" (in a dictionary sense) and 1 otherwise.
Why not use == there?:
if s == "help"
help()
elsif s == "add"
puts"Enter 2 numbers"
a = STDIN.gets
b = STDIN.gets
add(a,b)
else
err("Invalid command, type 'help' for list of available commands")
end
Or to clean it up a bit:
case s
when "help":
help()
when "add":
puts"Enter 2 numbers"
a = STDIN.gets
b = STDIN.gets
add(a,b)
else
err("Invalid command, type 'help' for list of available commands")
end
Case is perfectly communicating the intent there.
Comparison operator(<=>) on String: Returns -1, 0, +1 or nil depending on whether a string is less than, equal to, or greater than other_string. nil is returned if the two values are incomparable.
All of the returning values (-1, 0, +1) are truthy in ruby. So what you wrote is equal to
if true
help ()
else
....
end
So the goal here is to print the index of the element if the element is in the array or print -1 if the element is not in the array. I have to do this using loops. PLEASE HELP!
def element_index(element, my_array)
while my_array.map.include? element do
puts my_array.index(element)
break
end
until my_array.include? element do
puts -1
break
end
end
p element_index("c", ["a","b","c"])
If it's OK to use Array#index, then
def element_index(elem, collection)
collection.index(elem) || -1
end
Or if it's a homework that you should not use Array#index, or you want to do this on arbitrary collections, then
def element_index(elem, collection)
collection.each_with_index.reduce(-1) do |default, (curr, index)|
curr == elem ? (return index) : default
end
end
By the way, I always turn to Enumerable#reduce when I want to iterate over a collection (array, map, set, ...) to compute one value.
This is an easy way but maybe it doesn't meet the criteria for "using loops":
def element_index(x, arr)
arr.index(x) || -1
end
element_index("c", ["a","b","c"]) #=> 2
element_index("d", ["a","b","c"]) #=> -1
To explicitly use a loop:
def element_index(x, arr)
arr.each_index.find { |i| arr[i] == x } || -1
end
As pointed out in the comments, we could instead write
arr.each_index.find(->{-1}) { |i| arr[i] == x }
element_index("c", ["a","b","c"]) #=> 2
element_index("d", ["a","b","c"]) #=> -1
I know this is an assignment, but I'll first cover this as if it were real code because it's teaching you some not-so-great Ruby.
Ruby has a method for doing this, Array#index. It returns the index of the first matching element (there can be more than one), or nil.
p ["a","b","c"].index("c") # 2
p ["a","b","c"].index("d") # nil
Returning -1 is inadvisable. nil is a safer "this thing does not exist" value because its never a valid value, always false (-1 and 0 are true in Ruby), and does not compare equal to anything but itself. Returning -1 indicates whomever came up with this exercise is converting it from another language like C.
If you must, a simple wrapper will do.
def element_index(element, array)
idx = array.index(element)
if idx == nil
return -1
else
return idx
end
end
I have to do this using loops.
Ok, it's homework. Let's rewrite Array#index.
The basic idea is to loop through each element until you find one which matches. Iterating through each element of an array is done with Array#each, but you need each index, that's done with Array#each_index. The element can be then gotten with array[idx].
def index(array, want)
# Run the block for each index of the array.
# idx will be assigned the index: 0, 1, 2, ...
array.each_index { |idx|
# If it's the element we want, return the index immediately.
# No need to spend more time searching.
if array[idx] == want
return idx
end
}
# Otherwise return -1.
# nil is better, but the assignment wants -1.
return -1
end
# It's better to put the thing you're working on first,
# and the thing you're looking for second.
# Its like verb( subject, object ) or subject.verb(object) if this were a method.
p index(["a","b","c"], "c")
p index(["a","b","c"], "d")
Get used to using list.each { |thing| ... }, that's how you loop in Ruby, along with many other similar methods. There's little call for while and for loops in Ruby. Instead, you ask the object to loop and tell it what to do with each thing. It's very powerful.
I have to do this using loops.
You approach is very creative. You have re-created an if statement using a while loop:
while expression do
# ...
break
end
Is equivalent to:
if expression
# ...
end
With expression being something like array.include? element.
How can I do the opposite?
To invert a (boolean) expression, you just prepend !:
if !expression
# ...
end
Applied to your while-hack:
while !expression do
# ...
break
end
The whole method would look like this:
def element_index(element, my_array)
while my_array.include? element do
puts my_array.index(element)
break
end
while !my_array.include? element do
puts -1
break
end
end
element_index("c", ["a","b","c"])
# prints 2
element_index("d", ["a","b","c"])
# prints -1
As I said at the beginning, this approach is very "creative". You are probably supposed to find the index using a loop (see Schwern's answer) instead of calling the built-in index.
I would like to make a program that checks to see if the number you enter is an even number. Sort of like making a leap year program but for any number divisible by 2.
Something along the lines of:
num = gets.chomp
while num != 0
if (num%2) == 0
puts 'yess'
else
puts 'nooo'
end
end
I knows there's something easy that I need to change for it to run.
(btw I just started learning Ruby yesterday!)
There are two problems here.
First being something that others have put, you need to make sure you turn the input into an integer using ".to_i" on your num variable.
Secondly, this code puts you into an infinite loop since you are using a "while" loop.
Since the number is only input once, you get stuck in the "while" loop forever no matter what the input is. Basically, "num" never stops being not 0.
You'd be better off using an if..else statement. Something like:
num = gets.chomp.to_i
if num != 0
if (num%2) == 0
puts 'yess'
else
puts 'nooo'
end
else
puts "that's 0, dude"
end
Integers have two methods for this. They are even? and odd?.
You can use this in your if statement as so:
if num.even?
puts 'yess'
else
puts 'nooo'
end
However, an easier way to write this is with ternary expressions:
puts num.even? ? "yes" : "no"
However, make sure num is an Integer. Anything coming from gets will be a String. So, you should be doing num = gets.chomp.to_i. Anything that is not a number, like "h", will return 0.
"5".to_i #=> 5
"h".to_i #=> 0
Can you help me understand what this class does and how we can make use of it?
class Integer
def myt
c=0
until c == self
yield(c)
c+=1
end
self
end
end
Thank you.
x = Integer.new
x.myt
I tried to test it but it doesn't work. Error is: "no block given (yield)"
Also, in my book it says to test like this:
5.myt (|| puts "I'm on iteration #{i}! "} but it also gives an error - not sure why or what this line of code means.
allonhadaya and PNY did a good job explaining the purpose (enumeration) of the myt method.
Regarding your two questions mentioned in the title:
1.) What does 'c == self' do?
The '==' operator checks whether the integer c and Integer object you instantiate, are equal in value. If they are, the expression evaluates to true.
2.) What does 'yield' do?
The 'yield' statement passes control from the current method to a block which has been provided to the method. Blocks are ruby's implementation of a closure which, simple put, means that a method can be "extended" by calling the method with a block of additional code as long as the method supports a block (ie. incorporates yield statements)
The method seems to be a times implementation.
Basically 5.times { |i| puts i } and 5.myt { |i| puts i } will do exactly the same thing.
First, it sets a counter to 0, c = 0. Then you have a conditional where it checks if c is equal with self which will always be the integer attached to the method myt. It, then yields the counter and return self when is done.
Looks like it enumerates the values between zero inclusively and self exclusively.
allon#ahadaya:~$ irb
irb(main):001:0> class Integer
irb(main):002:1> def myt
irb(main):003:2> c=0
irb(main):004:2> until c == self
irb(main):005:3> yield(c)
irb(main):006:3> c+=1
irb(main):007:3> end
irb(main):008:2> self
irb(main):009:2> end
irb(main):010:1> end
=> nil
irb(main):011:0> 5.myt { |i| puts i }
0
1
2
3
4
=> 5
irb(main):012:0>
Using the example your book gave --
5.myt {|i| puts "I'm on iteration #{i}! "}
#You were missing an object in the pipes and a curly bracket before the pipes (not parentheses)
Allows you to see the internal workings of your myt method. Initializing variable c with a value of 0 the method executes an until look until the condition "c == self" is satisfied. Self references the object, here 5, which the method is acting on.
Therefore ...
def myt
until c == 5 #Until this is true
yield(c) #Do this .. here yield will do whatever the block specified
c+=1 #Increment on each iteration the value of variable c by 1
end #closing the until loop
self #return self
end
The yield within the method passes control from your method to the parameter, a block, back to the method.
Yield therefore allows you to build methods which can have similar patterns but with block you customize it to do your particular need.
If instead of putting each number maybe all you want to do is put the odd integers between 0 and the integer you call the method on --
5.myt {|i| puts i if i.odd?} # returns I am odd: 1 and I am odd: 3
I would suggest that you write your own blocks here to see how yield works and how you can keep the same method but pass in different blocks and create different method outputs!
This code does not perform as I expect:
case
when -> { false } then "why?"
else "This is what I expect"
end
# => "why?"
Neither does this
case
when ->(x) {false} then "why?"
else "This is what I expect"
end
# => "why?"
The first then clause is executed in both cases, which must mean that the lambda I supply to the when clause is not being called. I understand that the case equality operator === should be called on whatever the subject of the when clause is. I am wondering what goes on the other side of the === when there is no argument supplied to case. I was thinking it might be nil, but it can't be:
-> {false} === nil
# => ArgumentError: wrong number of arguments (1 for 0)
->(x) {false} === nil
# => false
This performs as expected and, if it were being executed, would lead to my expected case results or an exception. Can someone explain the results above? It seems that the case equality operator isn't being used at all, and yet the first when clause is evaluating to true. BTW, I am doing this because the output of a case can be used for variable assignment and it is less wordy then having several elsif clauses. I would like to be able to use arbitrary Procs in a case statement with no argument.
case
when -> { false } then puts "why?"
else puts "This is what I expect"
end
case
when 'cat' then puts "why?"
else puts "This is what I expect"
end
case
when -> { false }.call then puts "why?"
else puts "This is what I expect"
end
outputs:
why?
why?
This is what I expect
As The Pickaxe ( http://pragprog.com/book/ruby3/programming-ruby-1-9 ) says, there are two forms of case statement.
The first allows a series of conditions to be evaluated, executing
code corresponding to the first condition that is true: case when ‹
boolean-expression ›+ ‹ then › ...
The second form of a case expression takes a target expression
following the case keyword. case target when ‹ comparison ›+ ‹ then ›
...
In your case (case without target) any expression that is not false or nil (such as a Proc or the string 'cat') evaluates to true. The Proc is not executed, unless you call it.
You can use a case statement without a parameter to act similar to an if statement. For example:
case
when x > 0 then puts "positive"
else puts "negative"
end
You're assuming that it's trying to compare to nil, which isn't the case. Rather, when there's no parameter, the case statement is only testing for a "truthy" value (anything except nil and false). So when it hits your first when statement, it's checking to see if your Proc (meaning the actual ruby Proc object, not the results of executing your Proc) is nil or false, which isn't. Since it's "truthy", that code gets executed.
Try this, and you'll notice the Proc never even gets called/executed (you'll only see "bar", not "foo"):
case
when -> { puts 'foo' } then puts 'bar'
else puts "This line will never be printed"
end