CODEWITHSUNDEEP
rubysyntax
Cameron Booth
Cameron Booth

Reputation: 6992

Syntax for putting a block on a single line

So I've got a Ruby method like this:

def something(variable, &block)
  ....
end

And I want to call it like this:

something 'hello' { do_it }

Except that isn't working for me, I'm getting a syntax error. If I do this instead, it works:

something 'hello' do
  do_it
end

Except there I'm kind of missing the nice look of it being on one line.

I can see why this is happening, as it could look like it's a hash being passed as a second variable, but without a comma in between the variables...but I assume that there must be a way to deal with this that I'm missing. Is there?

Upvotes: 33

Views: 15283

Answers (3)

Ryan Bigg
Ryan Bigg

Reputation: 107738

Uh, what about:

>> def something(arg1 , &block)
>>   yield block
>> end
=> nil
>> def do_it
>>   puts "Doing it!"
>> end
=> nil
>> something('hello') { do_it }
"Doing it!"
=> nil

Upvotes: 0

Rômulo Ceccon
Rômulo Ceccon

Reputation: 10367

You need to parenthesize your argument:

something('hello') { do_it }

That should work.

Upvotes: 51

seanbehan
seanbehan

Reputation: 1621

If you want "def something" to to accept a block, you need to yield data to that block. For example: 

#to uppercase string
def something(my_input)
 yield my_input.upcase
end

# => "HELLO WORLD"
something("hello world") { |i| puts i}

Upvotes: 2

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