```On 1/9/06, Kroeger, Simon (ext) <simon.kroeger.ext / siemens.com> wrote:
> Hi,
>
> first: you don't need the 'to_a'
Excellent!  Less typing, and more importantly, less to parse when reading.

> second:
> ---------------------------------------
> (0..5).inject{|x,y| print y, ' '}
> puts
>
> (1..5).inject(0){|x,y| print y, ' '}
> puts
> ---------------------------------------
> output:
>
> 1 2 3 4 5
> 1 2 3 4 5
>
> If you provide no parameter to inject it will give
> you the first item of the enumerable as x.
>

That is true when you are using that first item, but for the dice
roll, we do not want the value of x in our accumulator, we are just
using length of the range to control the number of rolls.

In fact, maybe my original version was not a bug after all.  Look at
these simplified statements:

irb(main):041:0> (0..2).inject{|x,y|x+2}
=> 4
irb(main):042:0> (1..2).inject(0){|x,y|x+2}
=> 4

I looks like the first example above walks through [0,1,2] and calls
the x + 2 code block two times:
first time x=0 and y = 1, returns 2
second time x = 2 (accumulated from last time) and y=2, returns 4

The second one takes [1,2] and also calls the x + 2 code block twice.
first time x=0 (this time from the initialization value), y = 1, returns 2
second time x=2 and y = 2, returns 4

Ruby doc says this is because the  "second form uses the first element
of the collection as a the initial value (and skips that element while
iterating)."

I think that answers Greg's question too.

> > No, the mis-understanding is all mine.  You are right, otherwise it
> > will include an extra roll.  It should be:
> > (1..self).to_a.inject(0){|x,y| x + rand(sides)}+self
> >
> > Further proof of the value of good unit tests. (Since (1..1).to_a
> > returns a single-member array, you need to provide inject with an
> > initial value to make it work.)
> >
> > Regards,
> >
> > Paul.
> >
> >
> > On 1/9/06, Gregory Seidman <gsslist+ruby / anthropohedron.net> wrote:
> >
> > > }     # roll the dice
> > > }      (0..self).to_a.inject{|x,y| x + rand(sides)}+self
> > >
> > > I think I may be misunderstanding something here. You use
> > 0..self, when I
> > > would think it would have to be either 0...self or 1..self
> > to get the right
> > > number of rolls. Am I off? In my solution I used 1..self in
> > much the same
> > > way.
> >
> >
>
>

```