From: Stefano Crocco <stefano.crocco@...>
Date: 2007-08-28T00:27:27+09:00
Subject: Re: What is happening with this Unary minus?

Alle luned狸 27 agosto 2007, Todd Burch ha scritto:
> LOOPER = 6
>
> -(LOOPER).upto(LOOPER) {|i|
> puts i }
>
> I get one line of output:  6
>
> However, I get 13 lines of output here:
>
> (-LOOPER).upto(LOOPER) {|i|
> puts i }
>
> What is happening with the unary minus on the first example?  I expected
> to get identical output.
>
> Todd

I'm not completely sure, but I think the difference arises because of operator 
precedence. The first expression is interpreted as

-(LOOPER.upto(LOOPER){|i| puts i})

Since the lower and upper bounds are equal, the iteration is performed only 
one time. The - is then applied to the return value of upto (the receiver, 
i.e LOOPER). Indeed, if you try your code in irb, you'll see that the value 
of the expression is -6.

In the second case, using brackets you tell the interpreter that the upto 
method should not be called on LOOPER, but on (-LOOPER), that is on -6.

I hope this helps

Stefano