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 am learning Ruby, and for no particular reason, I want to pass the yield object to the Integer#times method so that the yield code block is called a number of times. Here is how I can do it with a named code block:
def withNamedCodeBlock &b
3.times(&b)
end
withNamedCodeBlock {print "Go "}
#returns Go Go Go
Now, I want to do the same, but without named code blocks; I want to do it with by using the yield keyword. Here is how I tried and failed:
def withYield
3.times(&yield)
end
withYield {print "Go "}
#returns Go => #<Enumerator: 3:times>
#I expect it to return Go Go Go
I am still wrapping my head around the various ways to pass code blocks to methods, so additional information regarding that is appreciated.
I want to pass the yield object to the Integer#times method so that the yield code block is called a number of times
yield is not an object, nor is it a block. It's not even a method call. It's a keyword that yields control to a passed block. If you want to do anything else with the block (save it for later, pass it around, etc.), you must name it.
For sake of completeness on this topic, I wanted to demonstrate another technique to call the original block:
def withProc
p = Proc.new
3.times(&p)
end
withProc { print "Go" }
When Proc.new is not given a block, it uses the block that was given to withProc instead. Now you can call p, and it will call the original block. You can also pass p to other methods like times either as a regular argument or as a block argument.
See https://medium.com/#amliving/proc-new-trick-c1df16185599 for more discussion
#sergio-tulentsev's answer is good. But, I wanted to point out that you can wrap yield in a new block and thereby pass along the ability to yield to the original block:
def withYield
3.times { yield }
end
withYield {print "Go "}
To be clear, { yield } is a new block, and it is passed to times. When the new block is executed, it yields to the original { print "Go" } block that was given to withYield. The original block isn't actually passed to times, but the ability to yield to the original block is passed, effectively letting you call the original block.
Ruby provides the #tap method, which allows you to take in a variable and run code on it, but then return the original variable rather than the result of your expression, ie:
def number
5.tap { |x| print x } # Prints 5, and returns 5
end
Is there any function built into Clojure that can provide this functionality?
You're looking for doto. Here's your example, rewritten using it:
(doto 5
println)
It works similarly to the -> macro in that it passes the value through a series of functions. A key difference is that it returns the initial value you passed in, rather than what is returned by the final function.
Taken from Programming ruby 1.9 book:
def my_while(cond, &body)
while cond.call
body.call
end
end
a=0
my_while -> { a < 3 } do
puts a
a += 1
end
produces:
0
1
2
The method expects an explicit parameter cond, and this "condition" is assumed to be a lambda/proc (the assumption is made by relying on cond.call to succeed) and has to be passed to the method my_while explicitly. The & syntax captures a method's block (if present) in a variable by implicitly converting it to a Proc object (see 'The ampersand').
Blocks are not real objects in Ruby and thus have to be converted by using the ampersand syntax. Once the block is bound to a Proc, you can send the call message on it as on any other proc/lambda.
The -> syntax is short for lambda, which converts a block to a Proc object (explicitly). There is also a slight difference between using lambda and Proc.new. Again, the wikibook:
Actually, there are two slight differences between lambda and Proc.new.
First, argument checking. The Ruby documentation for lambda states: Equivalent to Proc.new, except the resulting Proc objects check the number of parameters passed when called.
Second, there is a difference in the way returns are handled from the Proc. A return from Proc.new returns from the enclosing method (acting just like a return from a block, more on this later):
def try_ret_procnew
ret = Proc.new { return "Baaam" }
ret.call
"This is not reached"
end
# prints "Baaam"
puts try_ret_procnew
While return from lambda acts more conventionally, returning to its caller:
def try_ret_lambda
ret = lambda { return "Baaam" }
ret.call
"This is printed"
end
# prints "This is printed"
puts try_ret_lambda
With this in light, I would recommend using lambda instead of Proc.new, unless the behavior of the latter is strictly required. In addition to being way cooler a whopping two characters shorter, its behavior is less surprising.
The piece -> { a < 3 } is a shortcut for a lambda term (which was introduced with ruby 1.9). This is the first parameter passed to your method (i.e. cond) while the block afterwards is assigned to body. The lambda is then executed inside your method via cond.call.
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.