I have this method where it gets an input from the user and it checks it against a while condition. if the user inputted anything that isnt a string or if the user inputted a character that was longer than 1 the method would prompt the user again for a valid input, basically adhering to the hangman rules. Heres the code
class Hangman
def initialize
dictionary = File.open('5desk.txt',"r")
line = dictionary.readlines
#word = line[rand(1..line.length)]
#length = #word.length
random = #word.length - rand(#word.length/2)
random.times do
#word[rand(#word.length)] = "_"
end
end
This method fails to function properly.
def get_input
puts #word
puts "Letter Please?"
#letter = gets.chomp
while !#letter.kind_of? String || #letter.length != 1
puts "Invalid input,try again!"
#letter = gets.chomp
end
end
end
Game = Hangman.new
Game.get_input
class Hangman
Stop right there! Why create a class considering that you would only create a single instance of it? There's no need for one. A few methods and one instance variable are sufficient.
Generate secret words randomly
I assume the file '5desk.txt' contains one secret words per line and you will be selecting one randomly. So begin by gulping the entire file into an array held by an instance variable (as opposed to reading the file line-by-line). I assume '5desk.txt1' contains the three words shown below.
#secret_words = File.readlines('5desk.txt', chomp: true)
#=> ["cat", "violin", "whoops"]
See the doc for the class method IO::readlines1,2. The option chomp: true removes the newline character from the end of each line.
This method closes the file after it has been read. (You used File::open. When doing so you need to close the file when you are finished with it: f = File.open(fname)...f.close.)
You need a method to randomly choose a secret_word.
def fetch_secret_word
#secret_words.sample
end
fetch_secret_word
#=> "violin"
See Array#sample. You could have instead used
#secret_words[rand(#secret_words.size)]
See Kernel#rand. The first and last words in #secret_words are #secret_words[0] and #secret_words[#secret_words.size-1]. Therefore, where you wrote
#word = line[rand(1..line.length)]
it should have been
#word = line[rand(0..line.length-1)]
which is the same as
#word = line[rand(line.length)]
Now let's create a method for playing the game, passing an argument that equals the maximum number of incorrect guesses the player has before losing.
def play_hangman(max_guesses)
First get a secret word:
secret_word = fetch_secret_word
Let us suppose that secret_word #=> "violin"
Initialize objects
Next, initialize the number of incorrect guesses and an image of the secret word:
incorrect_guesses = 0
secret_word_image = "-" * secret_word.size
#=> "------"
So we now have
def play_hangman(max_guesses)
secret_word = fetch_secret_word
incorrect_guesses = 0
secret_word_image = "-" * secret_word.size
Loop over guesses
Now we need to loop over the player's guesses. I suggest you use Kernel#loop, in conjuction with the keyword break for all your looping needs. (For now, forget about while and until, and never use for.) The first thing we will do in the loop is to obtain the guess of a letter from the player, which I'll do by calling a method:
loop do
guess = get_letter(secret_word_image)
...<to be completed>
end
def get_letter(secret_word_image)
loop do
puts secret_word_image
print "Gimme a letter: "
letter = gets.chomp.downcase
break letter if letter.match?(/[a-z]/)
puts "That's not a letter. Try again."
end
end
guess = secret_letter(secret_word_image)
#=> "b"
Here this method returns "b" (the guess) and displays:
------
Gimme a letter: &
That's not a letter. Try again.
------
Gimme a letter: 3
That's not a letter. Try again.
------
Gimme a letter: b
See if letter guessed is in secret word
Now we need to see which if any of the hidden letters equal letter. Again, let's make this a method3.
def hidden_letters(guess, secret_word, secret_word_image)
(0..secret_word.size-1).select do |i|
guess == secret_word[i] && secret_word_image[i] = '-'
end
end
Suppose guess #=> "i". Then:
idx = hidden_letters(guess, secret_word, secret_word_image)
#=> [1,4]
There are two "i"'s, at indices 1 and 4. Had there been no hidden letters "i" the method would have returned an empty array.
Before continuing let's look at our play_hangman is coming along.
def play_hangman(max_guesses)
secret_word = fetch_secret_word
incorrect_guesses = 0
secret_word_image = "-" * secret_word.size
loop do
unless secret_word_image.include?('-')
puts "You've won. The secret word is '#{secret_word}'!"
break
end
guess = get_letter(secret_word_image)
idx = hidden_letters(guess, secret_word, secret_word_image)
...<to be completed>
end
Process a guess
We now have to carry out one course of action if the array idx is empty and another if it is not.
case idx.size
when 0
puts "Sorry, no #{guess}'s"
incorrect_guesses += 1
if incorrect_guesses == max_guesses
puts "Oh, my, you've used up all your guesses, but"
puts "we'd like you take home a bar of soap"
break
else
puts idx.size == 1 ? "There is 1 #{guess}!" :
"There are #{idx} #{guess}'s!"
idx.each { |i| secret_word_image[i] = guess }
if secret_word_image == secret_word
puts "You've won!! The secret word is '#{secret_word}'!"
break
end
end
Complete method
So now let's look at the full method (which calls fetch_secret_word, get_letter and hidden_letters).
def play_hangman(max_guesses)
secret_word = fetch_secret_word
incorrect_guesses = 0
secret_word_image = "-" * secret_word.size
loop do
guess = get_letter(secret_word_image)
idx = hidden_letters(guess, secret_word, secret_word_image)
case idx.size
when 0
puts "Sorry, no #{guess}'s"
incorrect_guesses += 1
if incorrect_guesses == max_guesses
puts "Oh, my, you've used up all your guesses,\n" +
"but we'd like you take home a bar of soap"
return
end
else
puts idx.size == 1 ? "There is 1 #{guess}!" :
"There are #{idx.size} #{guess}'s!"
idx.each { |i| secret_word_image[i] = guess }
if secret_word_image == secret_word
puts "You've won!! The secret word is '#{secret_word}'!"
return
end
end
end
end
Play the game!
Here is a example play of the game.
play_hangman(4)
------
Gimme a letter: #
That's not a letter. Try again.
------
Gimme a letter: e
Sorry, no e's
------
Gimme a letter: o
There is 1 o!
--o---
Gimme a letter: i
There are 2 i's!
-io-i-
Gimme a letter: l
There is 1 l!
-ioli-
Gimme a letter: v
There is 1 v!
violi-
Gimme a letter: r
Sorry, no r's
violi-
Gimme a letter: s
Sorry, no s's
violi-
Gimme a letter: t
Sorry, no t's
Oh, my, you've used up all your guesses,
but we'd like you take home a bar of soap
1 The class File has no (class) method readlines. So how can we write File.readlines? It's because File is a subclass of IO (File.superclass #=> IO) and therefore inherits IO's methods. One commonly sees IO class methods invoked with File as the receiver.
2 Ruby's class methods are referenced mod::meth (e.g., Array::new), where mod is the name of a module (which may be a class) and meth is the method. Instance methods are referenced mod#meth (e.g., Array#join).
3 Some Rubyists prefer to write (0..secret_word.size-1) with three dots: (0...secret_word.size). I virtually never use three dots because I find it tends to create bugs. The one exception is when creating an infinite range that excludes the endpoint (e.g., 1.0...1.5).
How can I find the line of the beginning and end of a Ruby method given a ruby file?
Say for example:
1 class Home
2 def initialize(color)
3 #color = color
4 end
5 end
Given the file home.rb and the method name initialize I would like to receive (2,4) which are the beginning and end lines.
Finding the end is tricky. The best way I can think of is to use the parser gem. Basically you'll parse the Ruby code into an AST, then recursively traverse its nodes until you find a node with type :def whose first child is :initialize:
require "parser/current"
def recursive_find(node, &block)
return node if block.call(node)
return nil unless node.respond_to?(:children) && !node.children.empty?
node.children.each do |child_node|
found = recursive_find(child_node, &block)
return found if found
end
nil
end
src = <<END
class Home
def initialize(color)
#color = color
end
end
END
ast = Parser::CurrentRuby.parse(src)
found = recursive_find(ast) do |node|
node.respond_to?(:type) && node.type == :def && node.children[0] == :initialize
end
puts "Start: #{found.loc.first_line}"
puts "End: #{found.loc.last_line}"
# => Start: 2
# End: 4
P.S. I would have recommended the Ripper module from the standard library, but as far as I can tell there's no way to get the end line out of it.
Ruby has a source_location method which gives you the file and the beginning line:
class Home
def initialize(color)
#color = color
end
end
p Home.new(1).method(:initialize).source_location
# => ["test2.rb", 2]
To find the end, perhaps look for the next def or EOF.
Ruby source is nothing but a text file. You can use linux commands to find the method line number
grep -nrw 'def initialize' home.rb | grep -oE '[0-9]+'
I have assumed that the file contains the definition of at most one initialize method (though generalizing the method to search for others would not be difficult) and that the definition of that method contains no syntax errors. The latter assumption is probably required for any method to extract the correct line range.
The only tricky part is finding the line containing end that is the last line of the definition of the initialize method. I've used Kernel#eval to locate that line. Naturally caution must be exercised whenever that method is to be executed, though here eval is merely attempting to compile (not execute) a method.
Code
def get_start_end_offsets(fname)
start = nil
str = ''
File.foreach(fname).with_index do |line, i|
if start.nil?
next unless line.lstrip.start_with?('def initialize')
start = i
str << line.lstrip.insert(4,'_')
else
str << line
if line.strip == "end"
begin
rv = eval(str)
rescue SyntaxError
nil
end
return [start, i] unless rv.nil?
end
end
end
nil
end
Example
Suppose we are searching a file created as follows1.
str = <<-_
class C
def self.feline
"cat"
end
def initialize(arr)
#row_sums = arr.map do |row|
row.reduce do |t,x|
t+x
end
end
end
def speak(sound)
puts sound
end
end
_
FName = 'temp'
File.write(FName, str)
#=> 203
We first search for the line that begins (after stripping leading spaces) "def initialize". That is the line at index 4. The end that completes the definition of that method is at index 10. We therefore expect the method to return [4, 10].
Let's see if that's what we get.
p get_start_end_offsets(FName)
#=> [4, 10]
Explanation
The variable start equals the index of the line beginning def initialize (after removing leading whitespace). start is initially nil and remains nil until the "def initialize" line is found. start is then set to the index of that line.
We now look for a line line such that line.strip #=> "end". This may or may not be the end that terminates the method. To determine if it is we eval a string that contains all lines from the one that begins def initialize to the line equal to end just found. If eval raises a SyntaxError exception that end does not terminate the method. That exception is rescued and nil is returned. eval will return :_initialize (which is truthy) if that end terminates the method. In that case the method returns [start, i], where i is the index of that line. nil is returned if no initialize method is found in the file.
I've converted "initialize" to "_initialize" to suppress the warning (eval):1: warning: redefining Object#initialize may cause infinite loop)
See both answers to this SO question to understand why SyntaxError is being rescued.
Compare indentation
If it is known that "def initialize..." is always indented the same amount as the line "end" that terminates the method definition (and no other lines "end" between the two are indented the same), we can use that fact to obtain the beginning and ending lines. There are many ways to do that; I will use Ruby's somewhat obscure flip-flop operator. This approach will tolerate syntax errors.
def get_start_end_offsets(fname)
indent = -1
lines = File.foreach(fname).with_index.select do |line, i|
cond1 = line.lstrip.start_with?('def initialize')
indent = line.size - line.lstrip.size if cond1
cond2 = line.strip == "end" && line.size - line.lstrip.size == indent
cond1 .. cond2 ? true : false
end
return nil if lines.nil?
lines.map(&:last).minmax
end
get_start_end_offsets(FName)
#=> [4, 10]
1 The file need not contain only code.
this sounds weird doesn't it?
class Dummy
def foo=(value); end
end
Dummy.new.foo = 1 { |x| x } # => syntax error
Dummy.new.foo=(1) { |x| x } # => syntax error
i tried every permutation of blanks, parenthesis, commas; no luck. i'm puzzled. i never suspected methods ending with '=' were special. is it a bug? is it intended? if intended, why? is it documented? where? please share insight.
thanks
ps. ruby is 1.9.2p290 (2011-07-09 revision 32553) [x86_64-darwin11.0.1]
The syntax sugar for methods ending in = does make it special. You can still do things like pass multiple arguments to that method, or pass a block, but not in any pretty or convenient manner:
class Foo
def bar=(a,b=nil)
p [a,b]
if block_given?
yield "hi"
else
puts "No block"
end
end
end
f = Foo.new
f.bar = 42
#=> [42, nil]
#=> No block
f.bar = 42, 17
#=> [[42,17], nil]
#=> No block
f.send(:bar=,42,17) do |x|
puts "x is #{x.inspect}"
end
#=> [42, 17]
#=> x is "hi"
Another way in which these methods are special is that when called with the syntax sugar they evaluate to the right hand value, not the return value of the method:
class Foo
def bar=(a)
return 17 # really explicit
end
end
f = Foo.new
x = (f.bar = 42)
p x
#=> 42
x = f.send(:bar=,42)
p x
#=> 17
It's not so much that the methods themselves are special, but more to do with how Ruby deals with assignments (like foo = bar). First the right hand side is evalulated, then the left hand side is evaluated and the appropriate action is taken. If the left hand side is an object attribute, then the appropriate setter method is called.
So in your example:
Dummy.new.foo = 1 { |x| x }
First ruby tries to evalulate 1 { |x| x }, which is what causes the syntax error.
Dummy.new.foo=something doesn't actually mean "call the method named foo=", but actually means something more like "evalualate something, and then determine what `Dummy.new.foo is, and if it looks like an object attribute, add = to the name and call that method". This is why Dummy.new.foo= and Dummy.new.foo = both work the same way.
You can call these methods using send, and can pass a block with this:
Dummy.new.send "foo=", 2 do
puts "HI"
end
This is because with send you can explicitly name the method you want to call.
Of course the end result is that methods ending in = seem to have some "special" behaviour you need to be aware of, but it might be useful to understand what's actually going on.
What does the following code mean in Ruby?
||=
Does it have any meaning or reason for the syntax?
a ||= b is a conditional assignment operator. It means:
if a is undefined or falsey, then evaluate b and set a to the result.
Otherwise (if a is defined and evaluates to truthy), then b is not evaluated, and no assignment takes place.
For example:
a ||= nil # => nil
a ||= 0 # => 0
a ||= 2 # => 0
foo = false # => false
foo ||= true # => true
foo ||= false # => true
Confusingly, it looks similar to other assignment operators (such as +=), but behaves differently.
a += b translates to a = a + b
a ||= b roughly translates to a || a = b
It is a near-shorthand for a || a = b. The difference is that, when a is undefined, a || a = b would raise NameError, whereas a ||= b sets a to b. This distinction is unimportant if a and b are both local variables, but is significant if either is a getter/setter method of a class.
Further reading:
http://www.rubyinside.com/what-rubys-double-pipe-or-equals-really-does-5488.html
This question has been discussed so often on the Ruby mailing-lists and Ruby blogs that there are now even threads on the Ruby mailing-list whose only purpose is to collect links to all the other threads on the Ruby mailing-list that discuss this issue.
Here's one: The definitive list of ||= (OR Equal) threads and pages
If you really want to know what is going on, take a look at Section 11.4.2.3 "Abbreviated assignments" of the Ruby Language Draft Specification.
As a first approximation,
a ||= b
is equivalent to
a || a = b
and not equivalent to
a = a || b
However, that is only a first approximation, especially if a is undefined. The semantics also differ depending on whether it is a simple variable assignment, a method assignment or an indexing assignment:
a ||= b
a.c ||= b
a[c] ||= b
are all treated differently.
Concise and complete answer
a ||= b
evaluates the same way as each of the following lines
a || a = b
a ? a : a = b
if a then a else a = b end
-
On the other hand,
a = a || b
evaluates the same way as each of the following lines
a = a ? a : b
if a then a = a else a = b end
-
Edit: As AJedi32 pointed out in the comments, this only holds true if: 1. a is a defined variable. 2. Evaluating a one time and two times does not result in a difference in program or system state.
In short, a||=b means: If a is undefined, nil or false, assign b to a. Otherwise, keep a intact.
Basically,
x ||= y means
if x has any value leave it alone and do not change the value, otherwise
set x to y
It means or-equals to. It checks to see if the value on the left is defined, then use that. If it's not, use the value on the right. You can use it in Rails to cache instance variables in models.
A quick Rails-based example, where we create a function to fetch the currently logged in user:
class User > ActiveRecord::Base
def current_user
#current_user ||= User.find_by_id(session[:user_id])
end
end
It checks to see if the #current_user instance variable is set. If it is, it will return it, thereby saving a database call. If it's not set however, we make the call and then set the #current_user variable to that. It's a really simple caching technique but is great for when you're fetching the same instance variable across the application multiple times.
To be precise, a ||= b means "if a is undefined or falsy (false or nil), set a to b and evaluate to (i.e. return) b, otherwise evaluate to a".
Others often try to illustrate this by saying that a ||= b is equivalent to a || a = b or a = a || b. These equivalencies can be helpful for understanding the concept, but be aware that they are not accurate under all conditions. Allow me to explain:
a ||= b ⇔ a || a = b?
The behavior of these statements differs when a is an undefined local variable. In that case, a ||= b will set a to b (and evaluate to b), whereas a || a = b will raise NameError: undefined local variable or method 'a' for main:Object.
a ||= b ⇔ a = a || b?
The equivalency of these statements are often assumed, since a similar equivalence is true for other abbreviated assignment operators (i.e. +=,-=,*=,/=,%=,**=,&=,|=,^=,<<=, and >>=). However, for ||= the behavior of these statements may differ when a= is a method on an object and a is truthy. In that case, a ||= b will do nothing (other than evaluate to a), whereas a = a || b will call a=(a) on a's receiver. As others have pointed out, this can make a difference when calling a=a has side effects, such as adding keys to a hash.
a ||= b ⇔ a = b unless a??
The behavior of these statements differs only in what they evaluate to when a is truthy. In that case, a = b unless a will evaluate to nil (though a will still not be set, as expected), whereas a ||= b will evaluate to a.
a ||= b ⇔ defined?(a) ? (a || a = b) : (a = b)????
Still no. These statements can differ when a method_missing method exists which returns a truthy value for a. In this case, a ||= b will evaluate to whatever method_missing returns, and not attempt to set a, whereas defined?(a) ? (a || a = b) : (a = b) will set a to b and evaluate to b.
Okay, okay, so what is a ||= b equivalent to? Is there a way to express this in Ruby?
Well, assuming that I'm not overlooking anything, I believe a ||= b is functionally equivalent to... (drumroll)
begin
a = nil if false
a || a = b
end
Hold on! Isn't that just the first example with a noop before it? Well, not quite. Remember how I said before that a ||= b is only not equivalent to a || a = b when a is an undefined local variable? Well, a = nil if false ensures that a is never undefined, even though that line is never executed. Local variables in Ruby are lexically scoped.
If X does NOT have a value, it will be assigned the value of Y. Else, it will preserve it's original value, 5 in this example:
irb(main):020:0> x = 5
=> 5
irb(main):021:0> y = 10
=> 10
irb(main):022:0> x ||= y
=> 5
# Now set x to nil.
irb(main):025:0> x = nil
=> nil
irb(main):026:0> x ||= y
=> 10
x ||= y
is
x || x = y
"if x is false or undefined, then x point to y"
||= is a conditional assignment operator
x ||= y
is equivalent to
x = x || y
or alternatively
if defined?(x) and x
x = x
else
x = y
end
unless x
x = y
end
unless x has a value (it's not nil or false), set it equal to y
is equivalent to
x ||= y
Suppose a = 2 and b = 3
THEN, a ||= b will be resulted to a's value i.e. 2.
As when a evaluates to some value not resulted to false or nil.. That's why it ll not evaluate b's value.
Now Suppose a = nil and b = 3.
Then a ||= b will be resulted to 3 i.e. b's value.
As it first try to evaluates a's value which resulted to nil.. so it evaluated b's value.
The best example used in ror app is :
#To get currently logged in iser
def current_user
#current_user ||= User.find_by_id(session[:user_id])
end
# Make current_user available in templates as a helper
helper_method :current_user
Where, User.find_by_id(session[:user_id]) is fired if and only if #current_user is not initialized before.
||= is called a conditional assignment operator.
It basically works as = but with the exception that if a variable has already been assigned it will do nothing.
First example:
x ||= 10
Second example:
x = 20
x ||= 10
In the first example x is now equal to 10. However, in the second example x is already defined as 20. So the conditional operator has no effect. x is still 20 after running x ||= 10.
a ||= b
Signifies if any value is present in 'a' and you dont want to alter it the keep using that value, else if 'a' doesnt have any value, use value of 'b'.
Simple words, if left hand side if not null, point to existing value, else point to value at right side.
a ||= b
is equivalent to
a || a = b
and not
a = a || b
because of the situation where you define a hash with a default (the hash will return the default for any undefined keys)
a = Hash.new(true) #Which is: {}
if you use:
a[10] ||= 10 #same as a[10] || a[10] = 10
a is still:
{}
but when you write it like so:
a[10] = a[10] || 10
a becomes:
{10 => true}
because you've assigned the value of itself at key 10, which defaults to true, so now the hash is defined for the key 10, rather than never performing the assignment in the first place.
It's like lazy instantiation.
If the variable is already defined it will take that value instead of creating the value again.
Please also remember that ||= isn't an atomic operation and so, it isn't thread safe. As rule of thumb, don't use it for class methods.
This is the default assignment notation
for example: x ||= 1
this will check to see if x is nil or not. If x is indeed nil it will then assign it that new value (1 in our example)
more explicit:
if x == nil
x = 1
end
b = 5
a ||= b
This translates to:
a = a || b
which will be
a = nil || 5
so finally
a = 5
Now if you call this again:
a ||= b
a = a || b
a = 5 || 5
a = 5
b = 6
Now if you call this again:
a ||= b
a = a || b
a = 5 || 6
a = 5
If you observe, b value will not be assigned to a. a will still have 5.
Its a Memoization Pattern that is being used in Ruby to speed up accessors.
def users
#users ||= User.all
end
This basically translates to:
#users = #users || User.all
So you will make a call to database for the first time you call this method.
Future calls to this method will just return the value of #users instance variable.
As a common misconception, a ||= b is not equivalent to a = a || b, but it behaves like a || a = b.
But here comes a tricky case. If a is not defined, a || a = 42 raises NameError, while a ||= 42 returns 42. So, they don't seem to be equivalent expressions.
irb(main):001:0> a = 1
=> 1
irb(main):002:0> a ||= 2
=> 1
Because a was already set to 1
irb(main):003:0> a = nil
=> nil
irb(main):004:0> a ||= 2
=> 2
Because a was nil
This ruby-lang syntax. The correct answer is to check the ruby-lang documentation. All other explanations obfuscate.
Google
"ruby-lang docs Abbreviated Assignment".
Ruby-lang docs
https://docs.ruby-lang.org/en/2.4.0/syntax/assignment_rdoc.html#label-Abbreviated+Assignment
a ||= b is the same as saying a = b if a.nil? or a = b unless a
But do all 3 options show the same performance? With Ruby 2.5.1 this
1000000.times do
a ||= 1
a ||= 1
a ||= 1
a ||= 1
a ||= 1
a ||= 1
a ||= 1
a ||= 1
a ||= 1
a ||= 1
end
takes 0.099 Seconds on my PC, while
1000000.times do
a = 1 unless a
a = 1 unless a
a = 1 unless a
a = 1 unless a
a = 1 unless a
a = 1 unless a
a = 1 unless a
a = 1 unless a
a = 1 unless a
a = 1 unless a
end
takes 0.062 Seconds. That's almost 40% faster.
and then we also have:
1000000.times do
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
a = 1 if a.nil?
end
which takes 0.166 Seconds.
Not that this will make a significant performance impact in general, but if you do need that last bit of optimization, then consider this result.
By the way: a = 1 unless a is easier to read for the novice, it is self-explanatory.
Note 1: reason for repeating the assignment line multiple times is to reduce the overhead of the loop on the time measured.
Note 2: The results are similar if I do a=nil nil before each assignment.