Why does the break statement in ruby behave differently when using Proc.new v. the ampersand sign?

Question

The break statement for blocks (as per The Ruby Programming Language) is defined as follows:

it causes the block to return to its iterator and the iterator to return to the method that invoked it.

Therefore when the following code is run, it results in a LocalJumpError.

def test
    puts "entering test method"
    proc = Proc.new { puts "entering proc"; break }
    proc.call # LocalJumpError: iterator has already returned
    puts "exiting test method"
end
test

While the following code does not throw a LocalJumpError. What is special about the ampersand sign? Doesn't the ampersand sign implicitly use Proc.new?

def iterator(&proc)
    puts "entering iterator"
    proc.call # invoke the proc
    puts "exiting iterator" # Never executed if the proc breaks
end

def test
    iterator { puts "entering proc"; break }
end
test

In other words, I read the ampersand sign as a means of in-lining the Proc.new call. At which point the behavior should be just the same as the first code snippet.

def iterator (p = Proc.new { puts "entering proc"; break})
...
end

Disclaimer: I am newb learning the language (ruby 1.9.2), and therefore will appreciate references and a detailed synopsis.

Answer 1

break makes the block and the caller of the block return. In the following code:

proc = Proc.new { break }

The "caller" of the block which is converted to a Proc object is Proc.new. break is supposed to make the caller of the block return, but Proc.new has already returned.

In this code:

def iterator(&b); b.call; end
iterator { break }

The caller of the block is iterator, so it makes iterator return.