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comp.lang.ruby

Is faster using abbreviated parameter names?

Iñaki Baz Castillo

3/2/2009 8:49:00 PM

Hi, a simple question:

a)
def hello(string)
# stuff with 'string'
end

b)
def hello(s)
# stuff with 's'
end


Will be b) faster since the parameter name contains less letters?

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

Robert Klemme

3/2/2009 8:51:00 PM

0

On 02.03.2009 21:48, Iñaki Baz Castillo wrote:
> Hi, a simple question:
>
> a)
> def hello(string)
> # stuff with 'string'
> end
>
> b)
> def hello(s)
> # stuff with 's'
> end
>
>
> Will be b) faster since the parameter name contains less letters?

I have no idea. I guess not. But you can easily verify for yourself.
Just write up a little program using Benchmark and you'll soon know.

Kind regards

robert

Iñaki Baz Castillo

3/2/2009 9:06:00 PM

0

El Lunes, 2 de Marzo de 2009, Robert Klemme escribi=F3:
> On 02.03.2009 21:48, I=F1aki Baz Castillo wrote:
> > Hi, a simple question:
> >
> > a)
> > def hello(string)
> > # stuff with 'string'
> > end
> >
> > b)
> > def hello(s)
> > # stuff with 's'
> > end
> >
> >
> > Will be b) faster since the parameter name contains less letters?
>
> I have no idea. I guess not. But you can easily verify for yourself.
> Just write up a little program using Benchmark and you'll soon know.

Yes, doing a benchmark the result is more or less the same (any other facto=
r=20
seems to be more important), but what I want to know is what should be the=
=20
response based on how Ruby works. Under my understanding Ruby needs to pars=
e=20
during runtime the variable name so a longer variable name would require mo=
re=20
time, am I wrong?

Thanks a lot.


=2D-=20
I=F1aki Baz Castillo

Denis Haskin

3/2/2009 9:07:00 PM

0

[Note: parts of this message were removed to make it a legal post.]

And how much faster would it have to be to make it worth it to use a
meaningless name versus a meaningful one?

Smells like premature optimization to me...

Cheers,

dwh


Robert Klemme wrote:
> On 02.03.2009 21:48, Iñaki Baz Castillo wrote:
>> Hi, a simple question:
>>
>> a)
>> def hello(string)
>> # stuff with 'string'
>> end
>>
>> b)
>> def hello(s)
>> # stuff with 's'
>> end
>>
>>
>> Will be b) faster since the parameter name contains less letters?
>

Brian Candler

3/2/2009 9:27:00 PM

0

Iñaki Baz Castillo wrote:
> Yes, doing a benchmark the result is more or less the same (any other
> factor
> seems to be more important), but what I want to know is what should be
> the
> response based on how Ruby works. Under my understanding Ruby needs to
> parse
> during runtime the variable name so a longer variable name would require
> more
> time, am I wrong?

I believe you're wrong.

Symbols are resolved into references to the symbol table at parse time,
so when running, :s and :ssssssssssssssss are just two different
pointers into the same symbol table. As for local variables, they are
just offsets into the stack frame.

So it might take a microscopically small amount of extra time for your
program to start up, reading a few extra bytes of source code, but once
it's running, each iteration should take the same time.
--
Posted via http://www.ruby-....

Iñaki Baz Castillo

3/2/2009 10:59:00 PM

0

El Lunes, 2 de Marzo de 2009, Brian Candler escribi=C3=B3:
> I=C3=B1aki Baz Castillo wrote:
> > Yes, doing a benchmark the result is more or less the same (any other
> > factor
> > seems to be more important), but what I want to know is what should be
> > the
> > response based on how Ruby works. Under my understanding Ruby needs to
> > parse
> > during runtime the variable name so a longer variable name would require
> > more
> > time, am I wrong?
>
> I believe you're wrong.
>
> Symbols are resolved into references to the symbol table at parse time,
> so when running, :s and :ssssssssssssssss are just two different
> pointers into the same symbol table. As for local variables, they are
> just offsets into the stack frame.
>
> So it might take a microscopically small amount of extra time for your
> program to start up, reading a few extra bytes of source code, but once
> it's running, each iteration should take the same time.

Thanks, that makes sense :)

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

Iñaki Baz Castillo

3/2/2009 11:00:00 PM

0

El Lunes, 2 de Marzo de 2009, Denis Haskin escribi=C3=B3:
> And how much faster would it have to be to make it worth it to use a
> meaningless name versus a meaningful one?
>
> Smells like premature optimization to me...

Sure, it was just curiosity :)
I will not code in Ruby to get a unreadable code like in other languages :)



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

Denis Haskin

3/3/2009 12:14:00 AM

0

[Note: parts of this message were removed to make it a legal post.]

Good to hear.

It's just remarkable how lots of people start at the end and work their
way backwards... ;-)

dwh


Iñaki Baz Castillo wrote:
> El Lunes, 2 de Marzo de 2009, Denis Haskin escribió:
>
>> And how much faster would it have to be to make it worth it to use a
>> meaningless name versus a meaningful one?
>>
>> Smells like premature optimization to me...
>>
>
> Sure, it was just curiosity :)
> I will not code in Ruby to get a unreadable code like in other languages :)
>
>
>
>