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comp.lang.ruby

Extrange behaviour using "if XXX ..." and "XXX if ..."

Iñaki Baz Castillo

3/29/2008 11:03:00 PM

Hi, note the difference in those both cases:

a)
if a=3D"QWEQWE" : a end
=3D> "QWEQWE"

b)
a if a=3D"QWEQWE"
NameError: undefined local variable or method `a' for main:Object


I know that the behaviour in a) is no recommended and maybe deprecated in a=
=20
future, but why it works in a) and not in b) ?


Thanks.



=2D-=20
I=C3=B1aki Baz Castillo
6 Answers

Tim Hunter

3/29/2008 11:22:00 PM

0

Iñaki Baz Castillo wrote:
> Hi, note the difference in those both cases:
>
> a)
> if a="QWEQWE" : a end
> => "QWEQWE"
>
> b)
> a if a="QWEQWE"
> NameError: undefined local variable or method `a' for main:Object
>
>
> I know that the behaviour in a) is no recommended and maybe deprecated in a
> future, but why it works in a) and not in b) ?

A method call without parentheses and a plain variable look the same, so
Ruby tries to determine between the two by looking for an assignment as
it parses your script. If it sees you assigning a value to "a", then it
knows that "a" is a variable, otherwise it assumes that "a" is a method
call. Parsing occurs from the top down and from left to right.

In case (a) the assignment to "a" occurs before any other reference, so
Ruby knows it's a variable.

In case (b), there has not yet been an assignment before the reference
to a, so Ruby assumes that "a" is a method call, yet there is no
definition for a so you get the message.


--
RMagick: http://rmagick.ruby...
RMagick 2: http://rmagick.ruby...rmagick2.html

Iñaki Baz Castillo

3/29/2008 11:42:00 PM

0

El Domingo, 30 de Marzo de 2008, Tim Hunter escribi=C3=B3:

> In case (b), there has not yet been an assignment before the reference
> to a, so Ruby assumes that "a" is a method call, yet there is no
> definition for a so you get the message.

b)
a if a=3D"QWEQWE"
NameError: undefined local variable or method `a' for main:Object


But why the first "a" is interpreted before the "if" stament? Imagine this=
=20
example:

defined?a
=3D> nil
a=3D"BLABLA" if 2 =3D=3D 3
=3D> nil
a
=3D> nil

After this code "a" will remain "nil" (or the previous value it had), so:
a=3D"BLABLA"
hasn't been interpreted since the "if" stament failed.

Also note this other example:
=20
defined?kk
=3D> nil
kk if nil
=3D> nil

Why in this last case the code doesn't return:
"undefined local variable or method kk' "
?

I assume the reason: "kk" is not interpreted if the condition fails.
In b) the first "a" is then interpreted AFTER the condition success, so in=
=20
that moment "a" has been already declared (into the condition), so it would=
=20
exist and not give an error.

Am I not right?



=2D-=20
I=C3=B1aki Baz Castillo

Tim Hunter

3/30/2008 12:06:00 AM

0

Iñaki Baz Castillo wrote:
> El Domingo, 30 de Marzo de 2008, Tim Hunter escribió:
>
>> In case (b), there has not yet been an assignment before the reference
>> to a, so Ruby assumes that "a" is a method call, yet there is no
>> definition for a so you get the message.
>
> b)
> a if a="QWEQWE"
> NameError: undefined local variable or method `a' for main:Object
>
>
> But why the first "a" is interpreted before the "if" stament? Imagine this
> example:
>
> defined?a
> => nil
> a="BLABLA" if 2 == 3
> => nil
> a
> => nil
>
> After this code "a" will remain "nil" (or the previous value it had), so:
> a="BLABLA"
> hasn't been interpreted since the "if" stament failed.
>
> Also note this other example:
>
> defined?kk
> => nil
> kk if nil
> => nil
>
> Why in this last case the code doesn't return:
> "undefined local variable or method kk' "
> ?
>
> I assume the reason: "kk" is not interpreted if the condition fails.
> In b) the first "a" is then interpreted AFTER the condition success, so in
> that moment "a" has been already declared (into the condition), so it would
> exist and not give an error.
>
> Am I not right?
>
>
>

You must distinguish between parsing and execution. Ruby parses your
program before Ruby executes your program.

Ruby decides how to interpret the reference to a during parsing, before
Ruby executes the if statements in your examples. Their success or
failure has nothing to do with the decision. In this case

a if a = "qweqwe"

Remember parsing occurs from left to right. Ruby sees the reference to a
to the left of the assignment to a. When Ruby first encounters a
reference to a it assumes that since it has not yet seen any assigment
to a, then a must be a method name.


--
RMagick: http://rmagick.ruby...
RMagick 2: http://rmagick.ruby...rmagick2.html

Iñaki Baz Castillo

3/30/2008 3:17:00 AM

0

El Domingo, 30 de Marzo de 2008, Tim Hunter escribi=C3=B3:

> You must distinguish between parsing and execution. Ruby parses your
> program before Ruby executes your program.
>
> Ruby decides how to interpret the reference to a during parsing, before
> Ruby executes the if statements in your examples. Their success or
> failure has nothing to do with the decision. In this case
>
> a if a =3D "qweqwe"
>
> Remember parsing occurs from left to right. Ruby sees the reference to a
> to the left of the assignment to a. When Ruby first encounters a
> reference to a it assumes that since it has not yet seen any assigment
> to a, then a must be a method name.

Ok, I understand. Thanks a lot.

=2D-=20
I=C3=B1aki Baz Castillo

matt

3/30/2008 4:47:00 AM

0

Iñaki Baz Castillo <ibc@aliax.net> wrote:

> El Domingo, 30 de Marzo de 2008, Tim Hunter escribió:
>
> > You must distinguish between parsing and execution. Ruby parses your
> > program before Ruby executes your program.
> >
> > Ruby decides how to interpret the reference to a during parsing, before
> > Ruby executes the if statements in your examples. Their success or
> > failure has nothing to do with the decision. In this case
> >
> > a if a = "qweqwe"
> >
> > Remember parsing occurs from left to right. Ruby sees the reference to a
> > to the left of the assignment to a. When Ruby first encounters a
> > reference to a it assumes that since it has not yet seen any assigment
> > to a, then a must be a method name.
>
> Ok, I understand. Thanks a lot.

I asked the same question a few months ago, and got some great answers:

<http://groups.google.com/group/comp.lang.ruby/browse_thread/t...
291e32ea2570/49ca1f19297bece6?lnk=st&q=#49ca1f19297bece6>

Understanding this point has been very useful to me subsequently...! m.

--
matt neuburg, phd = matt@tidbits.com, http://www.tidbits...
Leopard - http://www.takecontrolbooks.com/leopard-custom...
AppleScript - http://www.amazon.com/gp/product/...
Read TidBITS! It's free and smart. http://www.t...

Iñaki Baz Castillo

3/30/2008 9:27:00 PM

0

El Domingo, 30 de Marzo de 2008, matt neuburg escribi=F3:
> I asked the same question a few months ago, and got some great answers:
>
> <http://groups.google.com/group/comp.lang.ruby/browse_thread/t...
> 291e32ea2570/49ca1f19297bece6?lnk=3Dst&q=3D#49ca1f19297bece6>
>
> Understanding this point has been very useful to me subsequently...! m.

Thanks for the link matt,it's very useful to understand this "issue".

Regards.

=2D-=20
I=F1aki Baz Castillo