how it works with Proc object?

rubyonrails-talk
1

def test   yield end

test do   p "a" end

i don`t 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 don`t 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

3

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

yield() means: execute the block that is specified in the method call.

test do   p "a" end

That syntax calls test() and everything after 'do' is a block, which ruby automatically passes to the method.

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 `<main>'

Typically test() would be defined like this:

def test   if block_given?     yield   else #do something else     puts "goodbye"   end end

test()

--output:-- goodbye