Re: Anonymous scalar ref revisited

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Re^2: Anonymous scalar ref revisited by rovf (Priest) on Apr 06, 2010 at 11:46 UTC
Certainly a funny proposal (though it bails out if the subroutine itself is unnamed, but I admit that this is a border case which we will ignore). Still I'm puzzled how this can work. What is the `f` in this `\f` supposed to mean? It can't be a reference to the subroutine f, sind this would be `\&f` and wouldn't be useful either. Hence it must mean that we apply `\` to the result of a parameterless invocation of `f`, but this shouldn't work either. Which hat did you get this `f` out from? -- Ronald Fischer <ynnor@mm.st>	[reply] [d/l] [select]
Re^3: Anonymous scalar ref revisited by dk (Chaplain) on Apr 07, 2010 at 08:50 UTC
That works only because f() itself is so simple, and also because `my $ref=undef; $$ref=4711;` [download] works too. That's why it's not serious, but also probably answers your question more deeply - you don't need a special syntax for anonymous scalar-ref, it gets autovivified when necessary.	[reply] [d/l]