From: Michael Edgar <adgar@...>
Date: 2011-03-03T11:21:18+09:00
Subject: Re: What is inject doing here?

The first two elements of the range are the first two elements to be used as arguments to
the inject block:

First, s = 0, i = 1:

a = 1 % 2 == 0 ? 1 : -1; p a # a = -1

Then, s = -1, i = 2:

a = -1 % 2 == 0 ? 1 : -1; p a # a = 1

Then, s = 1, i = 3:

a = 3 % 2 == 0 ? 1 : -1; p a # a = -1

And the range is exhausted.

Michael Edgar
adgar@carboni.ca
http://carboni.ca/

On Mar 2, 2011, at 9:15 PM, Todd Benson wrote:

> Why does this not do what I expect?
> 
> irb(main):001:0> RUBY_VERSION
> => "1.9.2"
> irb(main):002:0> (0..3).inject {|s, i| a = i%2 == 0 ? 1 : -1; p a}
> -1
> 1
> -1
> => -1
> 
> I would think the result should be
> 
> 1
> -1
> 1
> -1
> => -1
> 
> Todd
>