How it works with Proc object?

Rails
1

def test
yield
end

test do
p “a”
end

i dont know how yield function does in detail. why did i write yield and then the p 'a' works? i know that it is a Proc object , but i dont know how it works. Is Proc do as a parameter?
test do
p “a”
end
and is this code which makes the output?
thanks a lot.

2

All Ruby methods have an invisible implicit block parameter. That
means that the signature of #test really looks like this:

def test(&block)
  # ...
end

When you write “yield”, you’re saying, “call the block that was passed
to this method”, so #test looks like this:

def test(&block)
  call &block and evaluate it
end

That means that if you call #test with { p “a” }, then #test’s
implementation is effectively

def test
  p "a"
end

~ jf

John F.
Principal Consultant, BitsBuilder
LI: http://www.linkedin.com/in/johnxf
SO: User John Feminella - Stack Overflow

3

On Sep 6, 1:47pm, John F. [email protected] wrote:

That means that if you call #test with { p “a” }, then #test’s
implementation is effectively

def test
 p "a"
end

With the difference that the block executes with access to the local
variables, value of self etc. that were present where the block was
defined

Fred

4

felix wrote in post #1020377:

def test
yield
end

In addition to John F.'s excellent explanation, note that your
definition of test() requires that a block be specified when calling
test():

def test
yield
end

test()

–output:–
ruby.rb:2:in test': no block given (yield) (LocalJumpError) from ruby.rb:5:in’

Typically test() would be defined like this:

def test
if block_given?
yield #i.e. execute the block
else #do something else
puts “goodbye”
end
end

test()

–output:–
goodbye

Also note, you can capture the block in a variable:

def test(&var_name) #& means put the block in this var
if block_given?
another_method(var_name)
else
puts “goodbye”
end
end

def another_method(x)
x.call #execute the block
end

test do
puts ‘hello’
end

–output:–
hello