# Re: Dice Roller (#61) We don't need no steenking leexer/pars **URL:** \ **Category:** Ruby\ **Created:** [January 9, 2006, 3:20pm UTC](https://www.ruby-forum.com/t/re-dice-roller-61-we-dont-need-no-steenking-leexer-pars/51942 "2006-01-09T15:20:48Z")\ **Posts on this page:** 2\ **Page:** 1
**Author:** ![KroegerS\_Simon\_Sext](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/k/57b2e6/32.png) [@KroegerS\_Simon\_Sext](https://www.ruby-forum.com/u/KroegerS_Simon_Sext)\ **Post date:** [January 9, 2006, 3:20pm UTC](https://www.ruby-forum.com/t/re-dice-roller-61-we-dont-need-no-steenking-leexer-pars/51942/1 "2006-01-09T15:20:48Z")
Hi, first: you don’t need the ‘to\_a’ ## 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. cheers Simon ---
**Author:** ![KroegerS\_Simon\_Sext](https://www.ruby-forum.com/letter_avatar_proxy/v4/letter/k/57b2e6/32.png) [@KroegerS\_Simon\_Sext](https://www.ruby-forum.com/u/KroegerS_Simon_Sext)\ **Post date:** [January 9, 2006, 5:03pm UTC](https://www.ruby-forum.com/t/re-dice-roller-61-we-dont-need-no-steenking-leexer-pars/51942/2 "2006-01-09T17:03:26Z")
On 1/9/06, Kroeger, Simon (ext) [removed\_email\_address@domain.invalid](mailto:removed_email_address@domain.invalid) wrote: > Hi, > > first: you don’t need the ‘to\_a’ > Excellent! Less typing, and more importantly, less to parse when > reading. > 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.