def mypr; if block_given?; then proc do yield; end; else proc do puts
“default”; end; end; end; p= mypr do puts “hi”; end; 5.times(&p); q= mypr;
5.times(&q);
Well, try this
def mypr; if block_given?; then proc do yield; end; else proc do
puts “default”; break; end; end; end; p= mypr do puts “hi”; break; end;
5.times(&p); q= mypr; 5.times(&q);
This is not me which have tried to obfuscate the code :-)))
…
Guy Decoux
-
Sorry for the one-liner.
-
Only the first break causes a problem:
def mypr; if block_given?; then proc do yield; end; else proc
do puts “default”; end; end; end; p= mypr do puts “hi”; break; end;
5.times(&p); q= mypr; 5.times(&q);
works as expected, that is, the “hi” block breaks out.
- The problem with the first break is that it is not in the lexical scope
of the times loop.
This just shows that ruby has a nice lexical semantics for break, unlike
perl in which the loop control statements act on the innermost dynamically
enclosing block, and if it has a label, it will search for it in the
dynamical scope.
I consider this an advantage. Look. In a perl code like
sub twice (&) {
$0;
$0;
}
… later …
BACK:
print “before twice:\n”;
twice { print “hello\n”; rand()<0.5 and goto BACK; };
Although you can jump to the label BACK, you must first make sure that the
sub `twice’ (which may be written by someone else) does not have a label
called BACK. Although ruby does not have labeled loop control statements and
gotos, it is still better if the loop control statements work in a lexical
fashion (that is, they will jump only with respect to the visible loop that
is near the jump-control statement in the code), especially because of the
greater role of iterators (as compared to perl).
This is of course only my opinion, you may still think that dynamic control
sturctures are better. And btw, ruby has both: exceptions are dynamic,
callcc and loop control are lexical in this sense.
- This has nothing to do with making a proc from the yield-block.
def mypr (&f); if block_given?; then f; else proc do puts “default”; end;
end; end; p= mypr do puts “hi”; break; end; 5.times(&p); q= mypr;
5.times(&q);
works the same and
def mypr (&f); if block_given?; then f; else proc do puts “default”; break;
end; end; end; p= mypr do puts “hi”; break; end; 5.times(&p); q= mypr;
5.times(&q);
fails the same as above.
ambrus
···
On Tue, Apr 20, 2004 at 12:01:44AM +0900, ts wrote:
