# A = b = c order of evaluation weird **URL:** \ **Category:** ruby-talk\ **Created:** [2 October 2007 09:54 UTC](https://rubytalk.org/t/a-b-c-order-of-evaluation-weird/41100 "2007-10-02T09:54:47Z")\ **Posts on this page:** 5\ **Page:** 2
**Author:** ![Arlen\_Christian\_Mar1](https://avatars.discourse-cdn.com/v4/letter/a/9de053/32.png) [@Arlen\_Christian\_Mar1](https://rubytalk.org/u/Arlen_Christian_Mar1)\ **Post date:** [2 October 2007 15:07 UTC](https://rubytalk.org/t/a-b-c-order-of-evaluation-weird/41100/21 "2007-10-02T15:07:59Z")
It demonstrates, at least, that one subscript op is evaluated left-to-right, thus we expect all of them to be so. > **···** > > On Tue, 2007-10-02 at 23:25 +0900, 7stud -- wrote: > > > Since your example only has one subscript > > expression, it doesn't shed any light on that issue. ---
**Author:** ![Yukihiro\_Matsumoto2](https://avatars.discourse-cdn.com/v4/letter/y/b9e5f3/32.png) [@Yukihiro\_Matsumoto2](https://rubytalk.org/u/Yukihiro_Matsumoto2)\ **Post date:** [3 October 2007 04:11 UTC](https://rubytalk.org/t/a-b-c-order-of-evaluation-weird/41100/22 "2007-10-03T04:11:29Z")
Hi, > **···** > > In message "Re: a = b = c order of evaluation weird" on Tue, 2 Oct 2007 23:25:20 +0900, 7stud -- \ writes: > > > I guess it's safer to say that the order of evaluating arguments is > > undefined, unless it is stated somewhere. > > > > Is it defined for ruby? > > Two different answers: > >   \* there's no official specification of the language, so that it is >     not defined at all. > >   \* but at least the author (me) strongly suggests left-to-right >     evaluation order, so it seems we can say it's "defined". > >               matz. ---
**Author:** ![7stud1](https://avatars.discourse-cdn.com/v4/letter/7/59ef9b/32.png) [@7stud1](https://rubytalk.org/u/7stud1)\ **Post date:** [2 October 2007 16:23 UTC](https://rubytalk.org/t/a-b-c-order-of-evaluation-weird/41100/23 "2007-10-02T16:23:47Z")
Arlen Christian Mart Cuss wrote: > **···** > > > On Tue, 2007-10-02 at 23:25 +0900, 7stud -- wrote: > > > > > Since your example only has one subscript > > > expression, it doesn't shed any light on that issue. > > > > It demonstrates, at least, that one subscript op is evaluated > > left-to-right, thus we expect all of them to be so. > > How do you know the subscript expression wasn't evaluated from right to > left? > -- > Posted via [http://www.ruby-forum.com/\](http://www.ruby-forum.com/%5C). ---
**Author:** ![Arlen\_Christian\_Mar1](https://avatars.discourse-cdn.com/v4/letter/a/9de053/32.png) [@Arlen\_Christian\_Mar1](https://rubytalk.org/u/Arlen_Christian_Mar1)\ **Post date:** [2 October 2007 17:00 UTC](https://rubytalk.org/t/a-b-c-order-of-evaluation-weird/41100/24 "2007-10-02T17:00:12Z")
> Arlen Christian Mart Cuss wrote: > \>\> Since your example only has one subscript > \>\> expression, it doesn't shed any light on that issue. > \> > \> It demonstrates, at least, that one subscript op is evaluated > \> left-to-right, thus we expect all of them to be so. > > How do you know the subscript expression wasn't evaluated from right to > left? We'll combine two pieces of information here: > But anyway, my take on this is that > >   a[0..a.size/2] = a[a.size\*2/3..-1] = nil > > is equivalent to: > >   a.=(0..a.size/2, a.=(a.size\*2/3..-1, nil)) And: > irb(main):001:0\> a = [1,2,3,4,5] > =\> [1, 2, 3, 4, 5] > irb(main):002:0\> b = 1 > =\> 1 > irb(main):003:0\> a[b] = (b = 10) > =\> 10 > irb(main):004:0\> a > =\> [1, 10, 3, 4, 5] The first one shows us that our initial double-assignment statement is decomposed into that there function call. Looking at the second one, we can also see that it's like this: a.=(b, b = 10) The second example shows us that `b' in the first parameter must therefore be evaluated first, before `b = 10' is. We can apply that to the original example and reason that "0..a.size/2" is evaluated before "a.=(a.size\*2/3..-1, nil)". So, you can conclude that the assignments must be evaluated left-to-right, because of they way they decompose into functions. Hope this helps your understanding. Arlen > **···** > > On Wed, 2007-10-03 at 01:23 +0900, 7stud -- wrote: > > > \> On Tue, 2007-10-02 at 23:25 +0900, 7stud -- wrote: ---
**Author:** ![F\_Senault](https://avatars.discourse-cdn.com/v4/letter/f/c37758/32.png) [@F\_Senault](https://rubytalk.org/u/F_Senault)\ **Post date:** [2 October 2007 18:25 UTC](https://rubytalk.org/t/a-b-c-order-of-evaluation-weird/41100/25 "2007-10-02T18:25:06Z")
You can play with set\_trace\_func : 20:16 fred@vodka:~/ruby\> cat toto.rb #! /usr/local/bin/ruby a = Array(1..100) set\_trace\_func proc { |e,f,l,i,b,c|   printf "%8s %s:%-2d %10s %8s\n",e,f,l,i,c if e == 'c-call' } a[0..a.size/2] \   = \   a[a.size\*2/3..-1] \   = \   nil 20:16 fred@vodka:~/ruby\> ruby toto.rb   c-call toto.rb:9 size Array   c-call toto.rb:9 / Fixnum   c-call toto.rb:11 size Array   c-call toto.rb:11 \* Fixnum   c-call toto.rb:11 / Fixnum   c-call toto.rb:12 = Array   c-call toto.rb:10 = Array (I didn't expect the real line numbers, by the way. Cooool !)    Fred > **···** > > Le 02 octobre à 18:23, 7stud -- a écrit : > > > Arlen Christian Mart Cuss wrote: > > > > > On Tue, 2007-10-02 at 23:25 +0900, 7stud -- wrote: > > > > > > > Since your example only has one subscript > > > > expression, it doesn't shed any light on that issue. > > > > > > It demonstrates, at least, that one subscript op is evaluated > > > left-to-right, thus we expect all of them to be so. > > > > How do you know the subscript expression wasn't evaluated from right to > > left? > > -- > Now that I think about it, having a (dense) group of Marketeers and > Salesweasels play russian roulette with, say, an 88 Flak would be better > than handing one of them an M1911 and convincing them that their odds > are better than with a revolver. (Matt Olson in ATSR) [Previous page](https://rubytalk.org/t/a-b-c-order-of-evaluation-weird/41100.md?page=1)