When I call the greeter method, it doesn't output the string to the console. I'm having a hard time understanding why. Can anyone help?
def greeter(name)
return "Hello #{name}!!"
end
def by_three?(num)
if num % 3 == 0
puts true
else
puts false
end
end
greeter("Michael")
by_three?(4)
return returns value from method, but doesn't print it. You need:
puts greeter("Michael")
Also you don't need this return, at all, method returns value of the last executed line.
Related
I found following code at the internet
class Test
def value
'string'
end
def inspect
'value'
end
end
def test(arg)
arg.tap { |i| i.value }
end
p test(Test.new)
Could anyone explain why it returns
p test(Test.new)
# >> value
Because arg.tap returns arg itself, and p prints arg.inspect, since you have overwritten theinspect method of Test, it returns a string 'value', so you the print result is value
see also:
- p vs puts in Ruby
- tap method
tap used for chain of methods. It runs the passed blocked and returns the input without any change.
It means
arg.tap { |i| i.value } will return arg
p method runs inspect method of passed object
I have this following function but somehow it doesn't get inside the if statement. This is my gist on line 68
I'm trying to evaluate if an object.country name has the same name as "example_of_a_country_name"
def find_population_of_country(liste_pays, country_name)
given_country_population = 0
liste_pays.each do |n|
if (n.country.eql?(country_name.upcase))
given_country_population = n.population
#I'm trying to see if it's output something here
puts country_name.upcase
end
return given_country_population
end
end
can you type this within your function and tell me what you get?:
def find_population_of_country(liste_pays, country_name)
liste_pays.select {|n| n.country.eql?(country_name.upcase) }
end
end
The problem is your line:
return given_country_population
This should be moved one line up, inside the if ... end block for the name comparison, otherwise, it's returning even for non-matches.
Here comes another Codecademy question:
The following challenge has been presented.
Define two methods in the editor:
A greeter method that takes a single string parameter, name, and
returns a string greeting that person. (Make sure to use return and
don't use print or puts.)
A by_three? method that takes a single integer parameter, number, and
returns true if that number is evenly divisible by three and false if
not. Remember, it's a Ruby best practice to end method names that
produce boolean values with a question mark.
The code I put in re: was..
def greeter(name)
return "Greet #{name}"
end
def by_three?(x)
if x % 3==0
returns true
else
return false
end
greeter("Brant")
by_three?(6)
The console then gives me the following error:
Did you define your greeter method?
It seems like I have. Am I wrong?
this would be it:
def greeter(name)
"Greet #{name}"
end
def by_three?(x)
x % 3 == 0
end
greeter("Brant") # => "Greet Brant"
by_three?(6) # => true
It looks like you did not add "end" after your else statement. Here you go.
#For the greeter method, i decided to use this format
def greeter(name)
return name
end
greeter("Hello Jane, good morning")
def by_three?(number)
if number % 3 != 1
return true
else
return false
end #Add end here to end your statement
end
by_three?(5)
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.
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 }