How do I pass a block with other arguments?

Question

def test(args,&block)
  yield
end

test 1, {puts "hello"}

Last line doesn't work. How do I pass a block with other arguments?

Answer 1

test(1){ puts "hello" }

or

test(1) do 
   puts "hello" 
end

or

blk = proc{ puts "hello" }
test(1, &blk)

You can check out this https://pine.fm/LearnToProgram/chap_10.html

As @Cary Swoveland suggested we can go slightly deeper.

Any Ruby method can implicitly accept a block. And even though you didn't define it in your method signature you still can capture it and pass further.

So, considering this idea we can do following manipulations with your method:

def test(args, &block)
  yield
end

is the same as

def test(args)
  yield
end

and the same as

def test(args)
   block = Proc.new
   block.call
end

When you have this implicit block capturing you'd probably want to add extra check:

def test(args)
   if block_given?
     block = Proc.new
     block.call
   else
     "no block"
   end
end

or

def test(args)
   if block_given?
     yield
   else
     "no block"
   end
end

So calling these methods will return following:

test("args")
#=> no block
test("args"){ "Hello World" }
#=> "Hello World"