Implicit return values in Ruby - ruby

I am somewhat new to Ruby and although I find it to be a very intuitive language I am having some difficulty understanding how implicit return values behave.
I am working on a small program to grep Tomcat logs and generate pipe-delimited CSV files from the pertinent data. Here is a simplified example that I'm using to generate the lines from a log entry.
class LineMatcher
class << self
def match(line, regex)
output = ""
line.scan(regex).each do |matched|
output << matched.join("|") << "\n"
end
return output
end
end
end
puts LineMatcher.match("00:00:13,207 06/18 INFO stateLogger - TerminationRequest[accountId=AccountId#66679198[accountNumber=0951714636005,srNumber=20]",
/^(\d{2}:\d{2}:\d{2},\d{3}).*?(\d{2}\/\d{2}).*?\[accountNumber=(\d*?),srNumber=(\d*?)\]/)
When I run this code I get back the following, which is what is expected when explicitly returning the value of output.
00:00:13,207|06/18|0951714636005|20
However, if I change LineMatcher to the following and don't explicitly return output:
class LineMatcher
class << self
def match(line, regex)
output = ""
line.scan(regex).each do |matched|
output << matched.join("|") << "\n"
end
end
end
end
Then I get the following result:
00:00:13,207
06/18
0951714636005
20
Obviously, this is not the desired outcome. It feels like I should be able to get rid of the output variable, but it's unclear where the return value is coming from. Also, any other suggestions/improvements for readability are welcome.

Any statement in ruby returns the value of the last evaluated expression.
You need to know the implementation and the behavior of the most used method in order to exactly know how your program will act.
#each returns the collection you iterated on. That said, the following code will return the value of line.scan(regexp).
line.scan(regex).each do |matched|
output << matched.join("|") << "\n"
end
If you want to return the result of the execution, you can use map, which works as each but returns the modified collection.
class LineMatcher
class << self
def match(line, regex)
line.scan(regex).map do |matched|
matched.join("|")
end.join("\n") # remember the final join
end
end
end
There are several useful methods you can use depending on your very specific case. In this one you might want to use inject unless the number of results returned by scan is high (working on arrays then merging them is more efficient than working on a single string).
class LineMatcher
class << self
def match(line, regex)
line.scan(regex).inject("") do |output, matched|
output << matched.join("|") << "\n"
end
end
end
end

In ruby the return value of a method is the value returned by the last statement. You can opt to have an explicit return too.
In your example, the first snippet returns the string output. The second snippet however returns the value returned by the each method (which is now the last stmt), which turns out to be an array of matches.
irb(main):014:0> "StackOverflow Meta".scan(/[aeiou]\w/).each do |match|
irb(main):015:1* s << match
irb(main):016:1> end
=> ["ac", "er", "ow", "et"]
Update: However that still doesn't explain your output on a single line. I think it's a formatting error, it should print each of the matches on a different line because that's how puts prints an array. A little code can explain it better than me..
irb(main):003:0> one_to_three = (1..3).to_a
=> [1, 2, 3]
irb(main):004:0> puts one_to_three
1
2
3
=> nil
Personally I find your method with the explicit return more readable (in this case)

Related

Appending to list in ruby

I am totally newbie in ruby, I want to create a list depending on the values of the stage, for this example. I have assigned the constant values. I am getting an empty array (NIL value).
PROD_WAVE1_STAGE = "prod-wave1"
PROD_WAVE2_STAGE = "prod-wave2"
PROD_WAVE3_STAGE = "prod-wave3"
def prod_dimensionality stage
whitelist = []
case stage
when 'prod-wave1'
whitelist << 'NRT'
when 'PROD_WAVE2_STAGE'
whitelist << 'SIN'
when 'PROD_WAVE3_STAGE'
whitelist << 'DUB'
when 'PROD_WAVE4_STAGE'
whitelist << 'IAD'
end
end
prod_dimensionality(PROD_WAVE1_STAGE)
While the program as written is not wrong, it is a bit dangerous with respect to maintenance. Remember that a case statement returns the value of the last expression to be executed. In your case, this is something like whitelist << 'SIN', and since Array#<< returns the array itself, you are returning whitelist, and this is exactly what you need.
But imagine that for reason of debugging, you would add additional statements - for instance a test print - to your program, so it looks like this:
def prod_dimensionality(stage)
...
case ....
end
puts "whitelist=#{whitelist}" # Debugging output
end
In this case, the program would return the result of the puts statement, which will be nil. The caller would not see the whitelist anymore.
Therefore it is safer to write a return expression explicitly:
def prod_dimensionality(stage)
...
case ....
end
whitelist # This is what will be returned
end
You could make a hash that has the mappings in it. Then use that to decide what gets mapped in
stage_mappings = { 'prod-wave1' => 'NRT', ... }
whitelist << stage_mappings[stage]
whitelist.compact # In case there's some nils in there :D

Syntax error, unexpected tIDENTIFIER, expecting ')' Ruby

I get the following error when running a simple method that takes in a proper noun string and returns the string properly capitalized.
def format_name(str)
parts = str.split
arr = []
parts.map do |part|
if part[0].upcase
else part[1..-1].downcase
arr << part
end
end
return arr.join(" ")
end
Test cases:
puts format_name("chase WILSON") # => "Chase Wilson"
puts format_name("brian CrAwFoRd scoTT") # => "Brian Crawford Scott"
The only possibility that the above code returns a blank output is because your arr is nil or blank. And the reason your arr is blank(yes it is blank in your case) because of this line of code:
if part[0].upcase
in which the statement would always return true, because with every iteration it would check if the first element of the part string can be upcased or not, which is true.
Hence, your else block never gets executed, even if this got executed this would have returned the same string as the input because you are just putting the plain part into the array arr without any formatting done.
There are some ways you can get the above code working. I'll put two cases:
# one where your map method could work
def format_name(str)
parts = str.split
arr = []
arr = parts.map do |part|
part.capitalize
end
return arr.join(" ")
end
# one where your loop code logic works
def format_name(str)
parts = str.split
arr = []
parts.map do |part|
arr << "#{part[0].upcase}#{part[1..-1].downcase}"
end
return arr.join(" ")
end
There are numerous other ways this could work. I'll also put the one I prefer if I am using just plain ruby:
def format_name(str)
str.split(' ').map(&:capitalize)
end
You could also read more about the Open Classes concept to put this into the String class of ruby
Also, checkout camelize method if you're using rails.

why return change variables while inside a class

I cannot understand this ruby behavior, the code explains better what I mean:
class DoNotUnderstand
def initialize
#tiny_array = [3,4]
test
end
def messing(ary)
return [ary[0]+=700, ary[1]+=999]
end
def test
puts #tiny_array.join(",") # before => 3,4
messing(#tiny_array)
puts #tiny_array.join(",") # after => 703,1003
end
end
question = DoNotUnderstand.new
#tiny_array was [3,4] and became [703,1003]
if I don't use a class, that happens:
#tiny = [1,2]
def messing(ary)
return [ary[0]+693,ary[1]+999]
end
puts #tiny.join(",") # before => 1,2
messing(#tiny)
puts #tiny.join(",") # after => 1,2
the array simply remains [1,2]
why?
The class is a red herring, and completely irrelevant to the issue.
In the first case, where the array was modified, you defined messing as:
def messing(ary)
return [ary[0]+=700, ary[1]+=999]
end
Whereas in the second case, where the array was not modified, you defined messing as:
def messing(ary)
return [ary[0]+693,ary[1]+999]
end
In one case you used +=, and in the other, you used merely +.
ary[0] += 700 is exactly equivalent to ary[0] = ary[0] + 700. In other words you are changing the value stored in the 0th index of ary.
In the second case you merely add to the values stored in the array and return the result, but in the first case you not only return the result, you also store it back in the array.
For an explanation of why modifying ary modifies #tiny_array, see this answer to the question Is Ruby pass by reference or by value?.
You're second code example (the one from outside the class) is missing the two characters in the first that make it work the way it does. In the first example, the += operator is used, modifying the array in place:
return [ary[0]+=700, ary[1]+=999]
In your second example, the + operator is used, leaving the array as is:
return [ary[0]+693,ary[1]+999]
If you change it use the += operator, it works the same way as the first code snippet.

Ruby puts command in modules returning nil

i am working though LearnTocodethehardway.com http://ruby.learncodethehardway.org/book/ex25.html
On ex25. In the example there is a module that has a bunch of methods that return or print values. The Print_last_word method when supplied with an array of strings just puts nil. it does this even in his example output. My question would then be why?
To be precise, it doesn't puts nil - it puts the last word and returns nil. Here's the example output:
>> Ex25.print_last_word(words)
wait. # <- this is the output
=> nil # <- this is the return value
puts always returns nil.
UPDATE
There seems to be a bug in print_first_word:
module Ex25
def Ex25.print_first_word(words)
word = words.pop(0)
puts word
end
end
Ex25.print_first_word(["foo", "bar", "baz"])
#=> nil
This is because ["foo", "bar", "baz"].pop(0) returns an empty array and puts [] just returns nil without printing anything.
A working implementation (in the exercise's style) could look like this:
module Ex25
def Ex25.print_first_word(words)
word = words.shift
puts word
end
end
To make it more easy to understand:
"puts" never is used for its return value. It is used for its side effect if you wanted a method that would return a word, you would then do something with that word.
words = ["apple", "cucumber", "apple"]
def give_me_first_word(array_of_words)
array_of_words.first
end
variable_to_be_used_later = give_me_first_word(words)
This function would return "apple"(and in this case the variable would be assigned "apple"), but it would puts nothing to the screen. This value would then be used in another program or function.
If you puts a value, you would then not need to return it to any other program as it's already served its purpose. It's actually intuitive because if puts also returned the value, it would be doing two things. The counterpart to "puts" would be "return" that simply returns the value and does not display it to the screen.

Use of yield and return in Ruby

Can anyone help me to figure out the the use of yield and return in Ruby. I'm a Ruby beginner, so simple examples are highly appreciated.
Thank you in advance!
The return statement works the same way that it works on other similar programming languages, it just returns from the method it is used on.
You can skip the call to return, since all methods in ruby always return the last statement. So you might find method like this:
def method
"hey there"
end
That's actually the same as doing something like:
def method
return "hey there"
end
The yield on the other hand, excecutes the block given as a parameter to the method. So you can have a method like this:
def method
puts "do somthing..."
yield
end
And then use it like this:
method do
puts "doing something"
end
The result of that, would be printing on screen the following 2 lines:
"do somthing..."
"doing something"
Hope that clears it up a bit. For more info on blocks, you can check out this link.
yield is used to call the block associated with the method. You do this by placing the block (basically just code in curly braces) after the method and its parameters, like so:
[1, 2, 3].each {|elem| puts elem}
return exits from the current method, and uses its "argument" as the return value, like so:
def hello
return :hello if some_test
puts "If it some_test returns false, then this message will be printed."
end
But note that you don't have to use the return keyword in any methods; Ruby will return the last statement evaluated if it encounters no returns. Thus these two are equivelent:
def explicit_return
# ...
return true
end
def implicit_return
# ...
true
end
Here's an example for yield:
# A simple iterator that operates on an array
def each_in(ary)
i = 0
until i >= ary.size
# Calls the block associated with this method and sends the arguments as block parameters.
# Automatically raises LocalJumpError if there is no block, so to make it safe, you can use block_given?
yield(ary[i])
i += 1
end
end
# Reverses an array
result = [] # This block is "tied" to the method
# | | |
# v v v
each_in([:duck, :duck, :duck, :GOOSE]) {|elem| result.insert(0, elem)}
result # => [:GOOSE, :duck, :duck, :duck]
And an example for return, which I will use to implement a method to see if a number is happy:
class Numeric
# Not the real meat of the program
def sum_of_squares
(to_s.split("").collect {|s| s.to_i ** 2}).inject(0) {|sum, i| sum + i}
end
def happy?(cache=[])
# If the number reaches 1, then it is happy.
return true if self == 1
# Can't be happy because we're starting to loop
return false if cache.include?(self)
# Ask the next number if it's happy, with self added to the list of seen numbers
# You don't actually need the return (it works without it); I just add it for symmetry
return sum_of_squares.happy?(cache << self)
end
end
24.happy? # => false
19.happy? # => true
2.happy? # => false
1.happy? # => true
# ... and so on ...
Hope this helps! :)
def cool
return yield
end
p cool {"yes!"}
The yield keyword instructs Ruby to execute the code in the block. In this example, the block returns the string "yes!". An explicit return statement was used in the cool() method, but this could have been implicit as well.

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