I'm up against something I don't quite understand.
Here's the class I've created, and the driver code I'm running to test my method:
class Dog < Array
def breathing?
self.length > 1
end
end
the_dog = Dog.new(["Arf", "Woof"])
puts the_dog.breathing?
# true
the_dog.shift
puts the_dog.breathing?
# false
How come, when I call #shift on the_dog, the_dog stops breathing?
Shift method returns the first element of self and removes it (shifting all other elements down by one). Returns nil if the array is empty.
a = [1,2,3]
a.shift
=> [2,3]
So, In your case the first argument of the array is removed. Once its removed, the condition the_dog.breathing? is returned false.
Related
I have the following two methods, which I believe should have the same behaviour disregarding their names:
def a=(*params)
params
end
def b(*params)
params
end
But when in fact I use them:
a=(1) # => 1
b(1) # => [1]
(a=1) == b(1) # => false
while interestingly:
(a=1,2) == b(1,2) # => true
Why isn't their behaviour the same?
Edit: forgot to wrap the above in a class / call with self. which accidentally produces the same behaviour but for a different reason. It has been pointed out in the answers.
It has nothing to do with splat. It's the assignment operator. In ruby, the assignment operator returns the value assigned. The return value from the method is ignored.
So a=1 return 1, not [1].
But, as mentioned by #mudasobwa, you're not even calling the method here. But if you were, that's what would happen (ignoring the return value).
class Foo
def a=(*params)
params
end
end
f = Foo.new
f.a = 1 # => 1
f.a = 1,2 # => [1, 2]
To not ignore the return value, call that setter without using assignment operator.
f.send 'a=', 1 # => [1]
The thing is that
a = 1
sets the local variable and does not call your method at all. Try with
def a=(*param)
puts "I AM HERE"
end
var= methods require an explicit receiver. To call your method, call it with an explicit receiver:
self.a = 1
It still won’t return anything but 1, because assignment methods return the value (the same way as initialize called through MyClass.new returns an instance, no matter what.) But you might check that splat works with:
def a=(*param)
puts param.inspect
end
self.a = 1
# [1]
#⇒ 1
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.
I don't understand how this isn't working. The program is supposed to take instance method second in the class Array and return the 2nd object in the array
class Array
def second(*arr)
arr.length <= 1 ? nil : arr[1]
end
end
#Test cases
Test.assert_equals(Array([1, 2, 3]), 2,) #Getting nil
Test.assert_equals(Array([]), nil) #passes
Test.assert_equals(Array([1]), nil) #passes
What am I doing wrong? if I remove class Array and test on second it works fine?
Why use *arr? If you're monkey-patching Array, then use self:
class Array
def second
self.length <= 1 ? nil : self[1]
end
end
p [1,2,3].second #=> 2
p [1].second #=> nil
p [].second #=> nil
In answer to what you're doing wrong, your code as written doesn't need the splat (*) operator (it also doesn't need to be patched into the Array class). While patching into Array and using self allows you to call it like [1,2].second, you could also write it as follows without patching into Array:
def second(arr)
arr.length <= 1 ? nil : arr[1]
end
Which would need to be called like second([1,2]).
To find out more about the splat operator *, try something like this explanation (I confess - the first Google result, but seems ok), but what it's doing in your example is turning your passed-in array into an array of an array - e.g. [1,2,3] becomes [[1,2,3]].
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!
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.