=== operator

Hi,

{} === Hash
=> false

case {}
when Hash then true
end
=> true

Hash === {}
=> true

Why is the first statement false when the second is true? I understand
case uses === to compare the objects? Perhaps it switches them round as
in the last statement?

Cheers,

Jon

Hi –

On Thu, 7 Jun 2007, Jon L. wrote:

Why is the first statement false when the second is true? I understand
case uses === to compare the objects? Perhaps it switches them round as
in the last statement?

Yes; === is called on the when expression(s), with the case object as
argument.

David

On 6/6/07, Jon L. [email protected] wrote:

Hi,

Maybe a more explicit notation helps

{} === Hash
{}.send(:===, Hash)
=> false

case {}
when Hash then true
when Hash.send(:===, {}) then true

end
=> true

Hash === {}
=> true

Why is the first statement false when the second is true? I understand
case uses === to compare the objects? Perhaps it switches them round as
in the last statement?
No the LHS is the receiver and the RHS is the first argument of the
message.
{}.send(:===, Hash) the === instance_method of Hash is called which is
inherited from Object and basically means equality, hence false.

Hash.send(:===, {}) the class method of Hash is called which is
inherited from Class and means is_a? hence true.

HTH
Robert

Am Donnerstag, 07. Jun 2007, 05:32:38 +0900 schrieb Robert D.:

On 6/6/07, [email protected] [email protected] wrote:

Yes; === is called on the when expression(s), with the case object as
argument.
Well you are right of course, I said now because I did not understand
what OP meant :frowning:

Some may say it’s an unexpected behaviour. Others will find
themselves detecting it as a welcome surprise. This is Ruby
at its best.

Bertram

On 6/6/07, [email protected] [email protected] wrote:

Hi –

Yes; === is called on the when expression(s), with the case object as
argument.
Well you are right of course, I said now because I did not understand
what OP meant :frowning:

Robert

Hi –

On Thu, 7 Jun 2007, Bertram S. wrote:

at its best.
I always thought it was just the logical way to do the case statement.
Since you’re testing something about the case object, you may not know
what it is or what its === method does:

case x # what is x?
when 1 …
when “yes” …
when C
when nil …
end

So I don’t think it would make sense to have a case construct where
=== was called on x.

David

On 07.06.2007 13:02, [email protected] wrote:

what OP meant :frowning:
when 1 …
when “yes” …
when C
when nil …
end

So I don’t think it would make sense to have a case construct where
=== was called on x.

Another reason why that would be an odd way to do it: all tests then
would have to be implemented in x’s class - now how much sense would
that make to do that? Just think of the type test (i.e. using class
objects in when clause)…

Kind regards

robert

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