How to break out from a ruby block?

Solution 1:

Use the keyword next. If you do not want to continue to the next item, use break.

When next is used within a block, it causes the block to exit immediately, returning control to the iterator method, which may then begin a new iteration by invoking the block again:

f.each do |line|              # Iterate over the lines in file f
  next if line[0,1] == "#"    # If this line is a comment, go to the next
  puts eval(line)
end

When used in a block, break transfers control out of the block, out of the iterator that invoked the block, and to the first expression following the invocation of the iterator:

f.each do |line|             # Iterate over the lines in file f
  break if line == "quit\n"  # If this break statement is executed...
  puts eval(line)
end
puts "Good bye"              # ...then control is transferred here

And finally, the usage of return in a block:

return always causes the enclosing method to return, regardless of how deeply nested within blocks it is (except in the case of lambdas):

def find(array, target)
  array.each_with_index do |element,index|
    return index if (element == target)  # return from find
  end
  nil  # If we didn't find the element, return nil
end

Solution 2:

I wanted to just be able to break out of a block - sort of like a forward goto, not really related to a loop. In fact, I want to break of of a block that is in a loop without terminating the loop. To do that, I made the block a one-iteration loop:

for b in 1..2 do
    puts b
    begin
        puts 'want this to run'
        break
        puts 'but not this'
    end while false
    puts 'also want this to run'
end

Hope this helps the next googler that lands here based on the subject line.

Solution 3:

If you want your block to return a useful value (e.g. when using #map, #inject, etc.), next and break also accept an argument.

Consider the following:

def contrived_example(numbers)
  numbers.inject(0) do |count, x|
    if x % 3 == 0
      count + 2
    elsif x.odd?
      count + 1
    else 
      count
    end
  end
end

The equivalent using next:

def contrived_example(numbers)
  numbers.inject(0) do |count, x|
    next count if x.even?
    next (count + 2) if x % 3 == 0
    count + 1
  end
end

Of course, you could always extract the logic needed into a method and call that from inside your block:

def contrived_example(numbers)
  numbers.inject(0) { |count, x| count + extracted_logic(x) }
end

def extracted_logic(x)
  return 0 if x.even?
  return 2 if x % 3 == 0
  1
end

Solution 4:

use the keyword break instead of return