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Raku
Chap. 6.4 substr example doesn't work
When I try it:
$ perl6 -e 'my $p = "perl 5"; substr($p, 6, 1, '6'); say $p'
Too many positional parameters passed; got 4 but expected between 2 and 3
in sub substr at src/gen/CORE.setting:2034
in block
Even in Perl 5 it doesn't work as expected :-) $ perl -E 'my $p = "perl 5"; substr($p, 6, 1, '6'); say $p' perl 56
Did you mean this for the Perl 5 equivalent?
perl -E 'my $p="perl5";substr($p,4,1,'6');say $p'
-----Original Message----- From: Klaus Brüssel [mailto:[email protected]] Sent: Tuesday, July 24, 2012 10:29 AM To: Christopher Bottoms Subject: [book] Chap. 6.4 substr example doesn't work (#73)
When I try it:
$ perl6 -e 'my $p = "perl 5"; substr($p, 6, 1, '6'); say $p'
Too many positional parameters passed; got 4 but expected between 2 and 3
in sub substr at src/gen/CORE.setting:2034
in block
Even in Perl 5 it doesn't work as expected :-) $ perl -E 'my $p = "perl 5"; substr($p, 6, 1, '6'); say $p' perl 56
Reply to this email directly or view it on GitHub: https://github.com/perl6/book/issues/73
I verified this.
I believe that the index starts at 0 not at 1, so the count should be 5, and not six. It seems that niecza correctly implements both, but rakudo only implements the 3 param version. Also, it seems that rakudo done not implement the lvalue version of the substr, and instead the spec has been changed to include substr-rw, which is the lvalue version of substr.
The section of the book should talk about these differences, or at least point to another part of the book that goes into detail.
