[#4766] Wiki — "Glen Stampoultzis" <trinexus@...>

21 messages 2000/09/04
[#4768] RE: Wiki — "NAKAMURA, Hiroshi" <nahi@...> 2000/09/04

Hi, Glen,

[#4783] Re: Wiki — Masatoshi SEKI <m_seki@...> 2000/09/04

[#4785] Re: Wiki — "NAKAMURA, Hiroshi" <nakahiro@...> 2000/09/05

Howdy,

[#4883] Re-binding a block — Dave Thomas <Dave@...>

16 messages 2000/09/12

[#4930] Perl 6 rumblings -- RFC 225 (v1) Data: Superpositions — Conrad Schneiker <schneik@...>

Hi,

11 messages 2000/09/15

[#4936] Ruby Book Eng. translation editor's questions — Jon Babcock <jon@...>

20 messages 2000/09/16

[#5045] Proposal: Add constants to Math — Robert Feldt <feldt@...>

15 messages 2000/09/21

[#5077] Crazy idea? infix method calls — hal9000@...

This is a generalization of the "in" operator idea which I

17 messages 2000/09/22

[#5157] Compile Problem with 1.6.1 — Scott Billings <aerogems@...>

When I try to compile Ruby 1.6.1, I get the following error:

15 messages 2000/09/27

[ruby-talk:5115] Re: Crazy idea? infix method calls

From: hal9000@...
Date: 2000-09-26 00:10:02 UTC
List: ruby-talk #5115
> |But I still think, as a special case, it woudl be a good idea if
> |    x in y
> |meant the same as
> |   y.include? x
> |
> |It is "pretty" (IMO) in the same way that the for loop is pretty...
>
> It's possible, because 'in' is the reserved word, so that `x in y' is
> not valid now.  Let us discuss about name, behavior, and
> least-surprise-ness. ;-)
>

Certainly let's discuss it then... ;)

I propose that the (pseudo) operator "in" (named as such) should
result in a call to include? "under the hood," i.e., x in y or
any arbitrary pair of objects x and y, should behave exactly the
same as y.include? x, returning true or false and making possible
syntax like: if node in binary_tree then ...

Now, is there any hidden catch, any unforeseen complication that
I am unaware of?

Since it mimics an operator, it will have to have a precedence.
That's all that comes to my mind. If I had had more sleep last
night, I might have an opinion about high versus low... :)
Presumably we could fix it so that for all operaors op1 and op2,
a op1 b in c op2 d  would mean the same as the expression
a op1 c.include?(b) op2 d.

Does anyone but me like this proposal at all?

Hal

--
Hal Fulton


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