From: Tanaka Akira <akr@...17n.org>
Date: 2002-08-13T17:09:48+09:00
Subject: [ruby-dev:17999] Re: Hash has default block?

In article <20020813075933.DBB611415@helium.ruby-lang.org>,
  "K.Kosako" <kosako@sofnec.co.jp> writes:

> $B$5$C$-5$IU$$$?$N$G$9$,!"(B
> Hash#dup$B$G!"(BHASH_PROC_DEFAULT$B%U%i%0$,%3%T!<$5$l$J$$$N$OJQ$@$H;W$$$^$9!#(B
> ($B$^$?$O(BHash#dup$B$G<jB3$-$N$[$&$r%/%j%"$9$k!)(B)

$B$*$)!#(B

$BB>$K$J$$$N$+$J$!!"$H;W$C$F(B replace $B$r;n$7$F$_$?$i$3$l$O$=$b$=$b(B
default $B$rCV$-49$($J$$$h$&$G$9$M!#(B
-- 
[$BEDCf(B $BE/(B][$B$?$J$+(B $B$"$-$i(B][Tanaka Akira]
$B!V$U$($m(B! $B$o$+$a$A$c$s:n@o$G$9$(C"=$B!W(B(Little Worker, $B7KM7@84](B)