On Mon, Dec 14, 2009 at 4:59 AM, Bertram Scharpf
<lists / bertram-scharpf.de> wrote:
> Hi,
>
> Am Montag, 14. Dez 2009, 16:05:54 +0900 schrieb Ruby Newbee:
>> Oh sorry I have found that.
>> need to convert the arguments to an array.
>>
>> "Hello, %s %s" %(["Matz!","again"])
>> => "Hello, Matz! again"
>
> Be aware that % is interpreted as an operator because of the
> string in front of it. There is also a shortcut
>
> ("Matz!","again")
>
> for
>
> Q("Matz!","again")
>
> which would be a string.

Actually, I'm not sure how that % before the ( is being seen by the parser.

By itself

 %("Matz!","again")
 => "\"Matz!\",\"again\""

For either Ruby 1.8.6 or 1.9.

%( should interpret everything up to the the matching ) as part of a
string including the "'s and the ,  I'm not sure why it doesn't do the
same thing as:

 > "Hello, %s %s" % "\"Matz!\",\"again\""
ArgumentError: too few arguments
	from (irb):3:in `%'
	from (irb):3

Since the format string needs two substitutions and we are only giving it one.

Perhaps a subtle Ruby parsing/lexing bug.

> What you do is applying the mod(%) operator to a string:
>
>   > %s %d %f" % [ "hi", 33, 0.618] #=> "hi 33 0.618000"

No, this is sending the message % to the string.  String#% is NOT mod,
the documentation (informally calls it format) and directs you to
Kernel#sprintf for further explanation.

Other that sharing the name :'&' with the methods in the various
Numeric subclasses, there's no meaning of mod.

-- 
Rick DeNatale

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