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comp.lang.ruby

Keyword parameters

Glenn

6/11/2008 7:22:00 PM

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

Hi everyone,

I have a question about keyword parameters in Ruby. I know that v1.8.6 doesn't do this and v1.9 does, and that you can 'trick' v1.8.6 into doing it with hashes.

My question is whether you can do this code:

def aProc(options = {})
a = options[:a] || 5
b = options[:b] || 6
c = options[:c] || 7
print a, b, c
end

in such a way where you wouldn't have to explicitly say that the local variable a is equal to the value in the hash with symbol :a? In other words, to do something like iterate through the hash and turn all of the symbol keys into local variable with their respective values?

Thanks,

Glenn


----- Original Message ----
From: Jason Roelofs <jameskilton@gmail.com>
To: ruby-talk ML <ruby-talk@ruby-lang.org>
Sent: Monday, March 3, 2008 4:36:49 PM
Subject: Re: Why is this not implemented in Ruby

Show me a language that does allow you to do this, I've never seen it.
Even then, Ruby doesn't deal with method overloads via parameter
lists, so there's a fundamental reason why it won't work:

def foo(a)
end

def foo(a, b)
end

def foo(a, b, c)
end

foo(1) # => ArgumentError: wrong number of arguments (1 for 3)


Ruby 1.9 has keyword arguments, so you'll be able to get past that
easily then. You can fake it in 1.8 with a hash:

def aProc(options = {})
a = options[:a] || 5
b = options[:b] || 6
c = options[:c] || 7
print a, b, c
end

aProc(:a => 1, :c => 3)

With Ruby 1.9 that will look like (I think, doing this from memory):

aProc(a: 1, c: 3)

Jason

On Mon, Mar 3, 2008 at 4:27 PM, Damjan Rems <d_rems@yahoo.com> wrote:
>
> It is probably a good reason but why is this not implemented in ruby.
>
> irb(main):005:0> def aProc(a=5,b=6,c=7)
> irb(main):006:1> print a,b,c
> irb(main):007:1> end
> irb(main):008:0> aProc
> 567=> nil
>
> irb(main):009:0> aProc(1,,3)
> SyntaxError: compile error
> (irb):9: syntax error, unexpected ','
> aProc(1,,3)
> ^
> from (irb):9
> from :0
>
> Why oh why I can not omit parameter in the middle of statement? I am
> forced to know its default value if I want to omit a parameter in the
> middle or at the begining of the statement.
>
>
> by
> TheR
> --
> Posted via http://www.ruby-....
>
>
2 Answers

Mikael Høilund

6/11/2008 10:28:00 PM

0


On Jun 11, 2008, at 21:21, Glenn wrote:

> [T]o do something like iterate through the hash and turn all of the =20=

> symbol keys into local variable with their respective values?


No.
What you *can* do, however, is to simply refer to each of these =20
options directly in the hash, making the hash values default instead. =20=

The na=EFve way:

options[:a] ||=3D 5
...

However, a more preferable way, however, is defining a hash with the =20
default, and then letting the options hash override these:

options =3D {:a =3D> 5, ...}.merge(options)

You can then access each option as option[:a] .... There are a few =20
caveat with this, but these aren't usually a problem.

--=20
Mikael H=F8ilund
http://ho...

fedzor

6/11/2008 11:03:00 PM

0

What makes an *actual* keyword argument, and why don't symbols suffice?


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