I'm working my way through a simple tutorial of each vs for loops in Ruby. This is supposed to be one of the simpler examples but for some reason, I don't understand the interaction between the yield statements and the for loop.
class MyEachThing
def each
yield 1
yield 42
yield 2
yield 42
yield 3
end
end
for i in MyEachThing.new
p i
end
# >> 1
# >> 42
# >> 2
# >> 42
# >> 3
Yield in this next example that I made up makes sense to me:
def calling
p yield(45)
end
calling {|i| i*2}
I just don't get how the first example works. Thank you for the help.
for i in MyEachThing.new
p i
end
is similar to this:
MyEachThing.new.each do |i|
p i
end
which means, you are calling each method on MyEachThing instance and passing i to the block.
And, yield is equivalent to: block.call means, you're calling the block with the passed argument (in this case i).
yield i is equivalent to: block.call(i) and your block is just printing the value of i.
Related
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 question already has answers here:
Closed 10 years ago.
Possible Duplicate:
What’s this &block in Ruby? And how does it get passes in a method here?
I dont Understand the &block part, what does it do?
here is an example:
def method_missing(method_name, *args, &block)
#messages << method_name
#object.send method_name, *args, &block
end
Blocks give you an opportunity to state a callback to pass on to a method.
The & is key here - like #pst mentioned, it "promotes" the block to a Proc and binds the Proc to the variable with the given name.
With &
def time(&block)
puts block
end
time
# => nil
time { foo }
# => #<Proc:0x00029bbc>
Without &
def time(block)
puts block
end
time { foo }
# => ArgumentError: wrong number of arguments (0 for 1)
# Because & isn't included, the method instead expected an arguement,
# but as a block isn't a arguement an error is returned.
Answering "And how would I pass it to another method?" comment by Brian:
Like this:
def compose init_value, n=2, &b
if n==0 then init_value else
b.call( compose init_value, n - 1, &b )
end
end
compose 2 do |n| n * n end
#=> 16
compose 2, 4 do |n| n * n end
#=> 65536
compose 2, 4 do |n| n * 0.5 end
#=> 0.125
This is a recursive method that recursively applies the same block to a number several times. Here, the block packaged into b argument gets called, but at the same time it is passed on recursively to compose method, while n argument is decremented by 1. In the same way, b could be passed to any method, like map, reduce, anything.
Whereas, should you not need to pass the block to another method, you could simply use yield:
def apply_block_to_1_2_3
return yield( 1 ), yield( 2 ), yield( 3 )
end
apply_block_to_1_2_3 { |n| n * n }
#=> [1, 4, 9]
May the force be with you.
It converts the block to a proc object that can be passed on to another method.
when you call a method with a block, there are 2 ways to use that block:
call yield inside method
convert it into a Proc object by prepending & to it
with second way you can pass it to another method.
so in your case it transforms the given block into a Proc and calling method_name with it.
think of it as you can pass a block just like any argument.
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.
(1..5).each do|x| puts yield(x) end do |x| return x*2 end
In my head this would loop 1 through 5 call the first block that would yield to the second block and put 2,4,6,8,10
Why does this not work and whats the easiest way to write this.
yield works within the methods. Quote from "Programming Ruby":
Within the method, the block may be invoked, almost as if it were a
method itself, using the yield statement.
So, if you want to make this code working, you can change it to something like this:
def f(n)
(1..n).each do |x|
puts yield(x)
end
end
f(5) do |x|
x * 2
end
If you don't want to define method you should put block into the variable and then use it:
b = Proc.new{|x| x *2 }
(1..5).each do |x|
puts b.call(x)
end
The easiest way to write this:
(1..5).each { |x| puts x*2 }
Let's say I have a function
def odd_or_even n
if n%2 == 0
return :even
else
return :odd
end
end
And I had a simple enumerable array
simple = [1,2,3,4,5]
And I ran it through map, with my function, using a do-end block:
simple.map do
|n| odd_or_even(n)
end
# => [:odd,:even,:odd,:even,:odd]
How could I do this without, say, defining the function in the first place? For example,
# does not work
simple.map do |n|
if n%2 == 0
return :even
else
return :odd
end
end
# Desired result:
# => [:odd,:even,:odd,:even,:odd]
is not valid ruby, and the compiler gets mad at me for even thinking about it. But how would I implement an equivalent sort of thing, that works?
edit
In reality, the solution to my problem matters to me a lot less than the motivation/reasoning behind it, to help me understand more how ruby blocks work :)
You're so close. Just remove the returns and you're golden.
This is because the block passed to map is a proc (i.e. created with Proc.new), and not a lambda. A return within a proc doesn't just jump out of the proc- it jumps out of the method that executed (i.e. called call on) the proc. A return within a lambda, on the other hand, jumps out of only the lambda.
The proc method returns a lambda in Ruby 1.8, and a Proc in Ruby 1.9. It's probably best to just not use this method and be explicit with which construct you want to use.
I'm guessing you were either in IRB or a plain ruby script when you were trying this out.
a = Proc.new { return }
a.call # fails. Nothing to return from.
def foobar
a = Proc.new { return }
a.call
puts 'hello' # not reached. The return within the proc causes execution to jump out of the foobar method.
end
foobar # succeeds, but does not print 'hello'. The return within the proc jumps out of the foobar method.
b = lambda { return }
b.call # succeeds. The return only returns from the lambda itself.
def bazquux
b = lambda { return }
b.call
puts 'hello' # this is reached. The lambda only returned from itself.
end
bazquux # succeeds, and prints 'hello'
The lesson to learn from this is to use implicit returns unless you can't, I guess.
I suspect this may be a duplicate question, but to give a value out of a block, use next
simple.map do |n|
if n%2 == 0
next :even
else
next :odd
end
end
Shortest variant using Andrew's answer:
simple.map { |n| next :even if n % 2 == 0; :odd }