There are many examples how to pass Ruby block as an argument, but these solutions pass the block itself.
I need a solution that takes some variable, executes an inline code block passing this variable as a parameter for the block, and the return value as an argument to the calling method. Something like:
a = 555
b = a.some_method { |value|
#Do some stuff with value
return result
}
or
a = 555
b = some_method(a) { |value|
#Do some stuff with value
return result
}
I could imagine a custom function:
class Object
def some_method(&block)
block.call(self)
end
end
or
def some_method(arg, &block)
block.call(arg)
end
but are there standard means present?
I think, you are looking for instance_eval.
Evaluates a string containing Ruby source code, or the given block, within the context of the receiver (obj). In order to set the context, the variable self is set to obj while the code is executing, giving the code access to obj’s instance variables. In the version of instance_eval that takes a String, the optional second and third parameters supply a filename and starting line number that are used when reporting compilation errors.
a = 55
a.instance_eval do |obj|
# some operation on the object and stored it to the
# variable and then returned it back
result = obj / 5 # last stament, and value of this expression will be
# returned which is 11
end # => 11
This is exactly how #Arup Rakshit commented. Use tap
def compute(x)
x + 1
end
compute(3).tap do |val|
logger.info(val)
end # => 4
Related
Example: LinkedList printing method.
For this object, you will find a printing method using block, proc, and lambda.
It is not clear to me what the advantages/disadvantages are (if any).
Thank you
What is a LinkedList?
A LinkedList is a node that has a specific value attached to it (which is sometimes called a payload), and a link to another node (or nil if there is no next item).
class LinkedListNode
attr_accessor :value, :next_node
def initialize(value, next_node = nil)
#value = value
#next_node = next_node
end
def method_print_values(list_node)
if list_node
print "#{list_node.value} --> "
method_print_values(list_node.next_node)
else
print "nil\n"
return
end
end
end
node1 = LinkedListNode.new(37)
node2 = LinkedListNode.new(99, node1)
node3 = LinkedListNode.new(12, node2)
#printing the linked list through a method defined within the scope of the class
node3.method_print_values(node3)
#---------------------------- Defining the printing method through a BLOCK
def block_print_value(list_node, &block)
if list_node
yield list_node
block_print_value(list_node.next_node, &block)
else
print "nil\n"
return
end
end
block_print_value(node3) { |list_node| print "#{list_node.value} --> " }
#---------------------------- Defining the printing method through a PROC
def proc_print_value(list_node, callback)
if list_node
callback.call(list_node) #this line invokes the print function defined below
proc_print_value(list_node.next_node, callback)
else
print "nil\n"
end
end
proc_print_value(node3, Proc.new {|list_node| print "#{list_node.value} --> "})
#---------------------------- Defining the printing method through a LAMBDA
def lambda_print_value(list_node, callback)
if list_node
callback.call(list_node) #this line invokes the print function defined below
lambda_print_value(list_node.next_node, callback)
else
print "nil\n"
end
end
lambda_print_value(node3, lambda {|list_node| print "#{list_node.value} --> "})
#---------------------------- Defining the printing method outside the class
def print_values(list_node)
if list_node
print "#{list_node.value} --> "
print_values(list_node.next_node)
else
print "nil\n"
return
end
end
print_values(node3)
Examples display how to use different things to do the same. So, there is no principal difference between them in this context:
my_proc = Proc.new { |list_node| print "#{list_node.value} --> " }
node3.block_print_values(node3, &my_proc)
node3.proc_print_value(node3, my_proc)
node3.lambda_print_value(node3, my_proc)
Also, there is possible to define a method by using any of them:
define_method(:my_method, p, &proc { puts p })
my_method 'hello' #=> hello
define_method(:my_method, p, &-> { puts p })
my_method 'hello' #=> hello
But Proc, Lambda, block are not the same. Firstly, need a bit more display how to works magic &. The great article can help with that:
&object is evaluated in the following way:
if object is a block, it converts the block into a simple proc.
if object is a Proc, it converts the object into a block while preserving the lambda? status of the object.
if object is not a Proc, it first calls #to_proc on the object and then converts it into a block.
But this does not show the differences between them. So, now let go to the ruby source:
Proc objects are blocks of code that have been bound to a set of local variables. Once bound, the code may be called in different contexts and still access those variables.
And
+lambda+, +proc+ and Proc.new preserve the tricks of a Proc object given by & argument.
lambda(&lambda {}).lambda? #=> true
proc(&lambda {}).lambda? #=> true
Proc.new(&lambda {}).lambda? #=> true
lambda(&proc {}).lambda? #=> false
proc(&proc {}).lambda? #=> false
Proc.new(&proc {}).lambda? #=> false
Proc created as:
VALUE block = proc_new(klass, FALSE);
rb_obj_call_init(block, argc, argv);
return block;
When lambda:
return proc_new(rb_cProc, TRUE);
Both are Proc. In this case, the difference is just in TRUE or FALSE. TRUE, FALSE - check the number of parameters passed when called.
So, lambda is like more strict Proc:
is_proc = !proc->is_lambda;
Summary of Lambda vs Proc:
Lambdas check the number of arguments, while procs do not.
Return within the proc would exit the method from where it is called.
Return within a lambda would exit it from the lambda and the method would continue executing.
Lambdas are closer to a method.
Blocks: They are called closures in other languages, it is a way of grouping code/statements. In ruby single line blocks are written in {} and multi-line blocks are represented using do..end.
Block is not an object and can not be saved in a variable. Lambda and Proc are both an object.
So, let do small code test based on this answer:
# ruby 2.5.1
user system total real
0.016815 0.000000 0.016815 ( 0.016823)
0.023170 0.000001 0.023171 ( 0.023186)
0.117713 0.000000 0.117713 ( 0.117775)
0.217361 0.000000 0.217361 ( 0.217388)
This shows that using block.call is almost 2x slower than using yield.
Thanks, #engineersmnky, for good references in comments.
Proc is an object wrapper over block. Lambda basically is a proc with different behavior.
AFAIK pure blocks are more rational to use compared to procs.
def f
yield 123
end
Should be faster than
def g(&block)
block.call(123)
end
But proc can be passed on further.
I guess you should find some articles with performance comparison on the toppic
IMO, your block_print_value method is poorly designed/named, which makes it impossible to answer your question directly. From the name of the method, we would expect that the method "prints" something, but the only printing is the border condition, which does a
print "nil\n"
So, while I would strongly vote against using this way to print the tree, it doesn't mean that the whole idea of using a block for the printing problem is bad.
Since your problem looks like a programming assignment, I don't post a whole solution, but give a hint:
Replace your block_print_value by a, say block_visit_value, which does the same like your current method, but doesn't do any printing. Instead, the "else" part could also invoke the block to let it do the printing.
I'm sure that you will see afterwards the advantage of this method. If not, come back here for a discussion.
At a high level, procs are methods that can be stored inside variables like so:
full_name = Proc.new { |first,last| first + " " + last }
I can call this in two ways, using the bracket syntax followed by the arguments I want to pass to it or use the call method to run the proc and pass in arguments inside of parentheses like so:
p full_name.call("Daniel","Cortes")
What I did with the first line above is create a new instance of Proc and assigned it to a variable called full_name. Procs can take a code block as a parameter so I passed it two different arguments, arguments go inside the pipes.
I can also make it print my name five times:
full_name = Proc.new { |first| first * 5 }
The block I was referring to is called a closure in other programming languages. Blocks allow you to group statements together and encapsulate behavior. You can create blocks with curly braces or do...end syntax.
Why use Procs?
The answer is Procs give you more flexibility than methods. With Procs you can store an entire set of processes inside a variable and then call the variable anywhere else in your program.
Similar to Procs, Lambdas allow you to store functions inside a variable and call the method from other parts of the program. So really the same code I had above can be used like so:
full_name = lambda { |first,last| first + " " + last }
p full_name["daniel","cortes"]
So what is the difference between the two?
There are two key differences in addition to syntax. Please note that the differences are subtle, even to the point that you may never even notice them while programming.
The first key difference is that Lambdas count the arguments you pass to them whereas Procs do not. For example:
full_name = lambda { |first,last| first + " " + last }
p full_name.call("Daniel","Cortes")
The code above works, however, if I pass it another argument:
p full_name.call("Daniel","Abram","Cortes")
The application throws an error saying that I am passing in the wrong number of arguments.
However, with Procs it will not throw an error. It simply looks at the first two arguments and ignores anything after that.
Secondly, Lambdas and Procs have different behavior when it comes to returning values from methods, for example:
def my_method
x = lambda { return }
x.call
p "Text within method"
end
If I run this method, it prints out Text within method. However, if we try the same exact implementation with a Proc:
def my_method
x = Proc.new { return }
x.call
p "Text within method"
end
This will return a nil value.
Why did this occur?
When the Proc saw the word return it exited out of the entire method and returned a nil value. However, in the case of the Lambda, it processed the remaining part of the method.
I found the following code here for eliminating duplicate records in an array:
require 'set'
class Array
def uniq_by
seen = Set.new
select{ |x| seen.add?( yield( x ) ) }
end
end
And we can use the code above as follows:
#messages = Messages.all.uniq_by { |h| h.body }
I would like to know how and what happens when the method is called. Can someone explain the internals of the code above? In the uniq_by method, we did not do anything to handle block argument. How is the passed argument handled by uniq_by method?
Let's break it down :
seen = Set.new
Create an empty set
select{ |x| seen.add?( yield( x ) ) }
Array#select will keep elements when the block yields true.
seen.add?(yield(x)) will return true if the result of the block can be added in the set, or false if it can't.
Indeed, yield(x) will call the block passed to the uniq_by method, and pass x as an argument.
In our case, since our block is { |h| h.body }, it would be the same as calling seen.add?(x.body)
Since a set is unique, calling add? when the element already exists will return false.
So it will try to call .body on each element of the array and add it in a set, keeping elements where the adding was possible.
The method uniq_by accepts a block argument. This allows to specify, by what criteria you wish to identify two elements as "unique".
The yield statement will evaluate the value of the given block for the element and return the value of the elements body attribute.
So, if you call unique_by like above, you are stating that the attribute body of the elements has to be unique for the element to be unique.
To answer the more specific question you have: yield will call the passed block {|h| h.body} like a method, substituting h for the current x and therefore return x.body
In Ruby, when you are putting yield keyword inside any method(say #bar), you are explicitly telling #bar that, you will be using a block with the method #bar. So yield knows, inside the method block will be converted to a Proc object, and yield have to call that Proc object.
Example :
def bar
yield
end
p bar { "hello" } # "hello"
p bar # bar': no block given (yield) (LocalJumpError)
In the uniq_by method, we did not do anything to handle block argument. How is the passed argument handled by uniq_by method?
You did do, that is you put yield. Once you will put this yield, now method is very smart to know, what it supposed to so. In the line Messages.all.uniq_by { |h| h.body } you are passing a block { |h| h.body }, and inside the method definition of uniq_by, that block has been converted to a Proc object, and yield does Proc#call.
Proof:
def bar
p block_given? # true
yield
end
bar { "hello" } # "hello"
Better for understanding :
class Array
def uniq_by
seen = Set.new
select{ |x| seen.add?( yield( x ) ) }
end
end
is same as
class Array
def uniq_by
seen = Set.new
# Below you are telling uniq_by, you will be using a block with it
# by using `yield`.
select{ |x| var = yield(x); seen.add?(var) }
end
end
Read the doc of yield
Called from inside a method body, yields control to the code block (if any) supplied as part of the method call. If no code block has been supplied, calling yield raises an exception. yield can take an argument; any values thus yielded are bound to the block's parameters. The value of a call to yield is the value of the executed code block.
Array#select returns a new array containing all elements of the array for which the given block returns a true value.
The block argument of the select use Set#add? to determine whether the element is already there. add? returns nil if there is already the same element in the set, otherwise it returns the set itself and add the element to the set.
The block again pass the argument (an element of the array) to another block (the block passed to the uniq_by) using yield; Return value of the yield is return value of the block ({|h| h.body })
The select .. statement is basically similar to following statement:
select{ |x| seen.add?(x.body) }
But by using yield, the code avoid hard-coding of .body, and defers decision to the block.
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.
I know that this code may be not quite correct:
def print_string(&str)
puts str
end
print_string{"Abder-Rahman"}
But, when I run it, this is what I get:
#<Proc:0x03e25d98#r.rb:5>
What is this output?
That's the default string representation of a Proc object. Because "Abder-Rahman" is in braces, Ruby thinks you're defining a block. Did you mean to put str.call inside of your function definition? That should call your block and return the string expression you defined inside it.
The problem is that you've declared that the "print_string" method takes a block argument (confusingly named "str") and you simply print the proc itself. You'd probably like to call the given procedure to see the string value it returns:
def call_proc(&proc)
proc.call
end
call_proc { 'Foobar' }
# => "Foobar"
What you've discovered is the syntax sugar that if you decorate the last argument of a method definition with an ampersand & then it will be bound to the block argument to the method call. An alternative way of accomplishing the same task is as follows:
def call_proc2
yield if block_given?
end
call_proc2 { 'Example' }
# => 'Example'
Note also that procedures can be handled directly as objects by using Proc objects (or the "lambda" alias for the Proc constructor):
p1 = Proc.new { 'Foo' }
p1.call # => "Foo"
p2 = lambda { 'Bar' }
p2.call # => "Bar"
You're passing a block to the method, as denoted by the & prefix and how you're calling it. That block is then converted into a Proc internally.
puts str.call inside your method would print the string, although why you'd want to define the method this way is another matter.
See Proc:
http://www.ruby-doc.org/core/classes/Proc.html
When the last argument of function/method is preceded by the & character, ruby expect a proc object. So that's why puts's output is what it is.
This blog has an article about the unary & operator.
In Ruby, is it possible to pass by reference a parameter with value-type semantics (e.g. a Fixnum)?
I'm looking for something similar to C#'s 'ref' keyword.
Example:
def func(x)
x += 1
end
a = 5
func(a) #this should be something like func(ref a)
puts a #should read '6'
Btw. I know I could just use:
a = func(a)
You can accomplish this by explicitly passing in the current binding:
def func(x, bdg)
eval "#{x} += 1", bdg
end
a = 5
func(:a, binding)
puts a # => 6
Ruby doesn't support "pass by reference" at all. Everything is an object and the references to those objects are always passed by value. Actually, in your example you are passing a copy of the reference to the Fixnum Object by value.
The problem with the your code is, that x += 1 doesn't modify the passed Fixnum Object but instead creates a completely new and independent object.
I think, Java programmers would call Fixnum objects immutable.
In Ruby you can't pass parameters by reference. For your example, you would have to return the new value and assign it to the variable a or create a new class that contains the value and pass an instance of this class around. Example:
class Container
attr_accessor :value
def initialize value
#value = value
end
end
def func(x)
x.value += 1
end
a = Container.new(5)
func(a)
puts a.value
You can try following trick:
def func(x)
x[0] += 1
end
a = [5]
func(a) #this should be something like func(ref a)
puts a[0] #should read '6'
http://ruby-doc.org/core-2.1.5/Fixnum.html
Fixnum objects have immediate value. This means that when they are assigned or
passed as parameters, the actual object is passed, rather than a reference to
that object.
Also Ruby is pass by value.
However, it seems that composite objects, like hashes, are passed by reference:
fp = {}
def changeit(par)
par[:abc] = 'cde'
end
changeit(fp)
p fp
gives
{:abc=>"cde"}