From: "David A. Black" <dblack@...>
Date: 2007-10-02T22:25:26+09:00
Subject: Re: a = b = c order of evaluation weird

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Hi --

On Tue, 2 Oct 2007, Mariusz P=C4~Ykala wrote:

> On 2007-10-02 21:06:32 +0900 (Tue, Oct), Calamitas wrote:
>> My impression was that the OP knew what the evaluation order is, but
>> not why it is like that.
>>
>> But anyway, my take on this is that
>>
>>   a[0..a.size/2] =3D a[a.size*2/3..-1] =3D nil
>>
>> is equivalent to:
>>
>>   a.[]=3D(0..a.size/2, a.[]=3D(a.size*2/3..-1, nil))
>>
>> Arguments are evaluated left to right, completely explaining the
>> evaluation order the OP is seeing.
>
> 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?

I believe that a subscript expression is always going to be evaluated
first:

irb(main):001:0> a =3D [1,2,3,4,5]
=3D> [1, 2, 3, 4, 5]
irb(main):002:0> b =3D 1
=3D> 1
irb(main):003:0> a[b] =3D (b =3D 10)
=3D> 10
irb(main):004:0> a
=3D> [1, 10, 3, 4, 5]


David

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