Re: Re: Re: Enforcing formal arguments at compile time?

I expect that in the example you give, foo will be called with a single scalar argument which is a reference to the array @ary. Am I wrong? Or isn't that what you expect?

Remember arrays are not lists. This is one of those occasions when forgetting that fact can leave you vaguely confused. Of course, if you've been following perl 6 you'd want to write that as foo(*@ary).

Update: Dang! I am wrong. I didn't expect it to be a compile error, but the direction Perl 6 is taking led me astray as to what would actually happen. Ah well.

Comment on Re: Re: Re: Enforcing formal arguments at compile time?

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Re: Re: Re: Re: Enforcing formal arguments at compile time? by tachyon (Chancellor) on May 03, 2002 at 12:07 UTC
Am I wrong? Yes, you are wrong. foo() is called with a single scalar argument which is precisely equal to `scalar @ary` or in other words the number of elements in @ary rather than a ref to it. I suggest you might like to 1) run the sample code where we print shift in foo and 2) follow the links I presented to learn more. I have a fair handle on scalar and array context BTW cheers tachyon s&&rsenoyhcatreve&&&s&n.+t&"$'$`$\"$\&"&ee&&y&srve&&d&&print	[reply]
Re:^5 Enforcing formal arguments at compile time? by pdcawley (Hermit) on May 03, 2002 at 15:52 UTC
Yick! I stand corrected. The arrayref version is `sub foo (\@) {...}` isn't it? I think I've been playing with Perl 6 a little too much recently...	[reply]