On 9/26/18 7:27 PM, Brandon Allbery wrote:
And again: this is only because you know perl 5. People are not born
knowing perl 5; to someone who doesn't know it, perldoc raises the same
kinds of questions you have been asking, and the answers have to be
found in perlsyn or perldata, etc. Which is exactly what you have been
complaining about with respect to perl 6 doing the same kind of thing.
Geez Louise Bradley! The above is a really bad argument!
"perldocs -f xxx" is a bazillion times easier to understand
than Perl 6's manual, regardless if you know Perl 5 or not.
And, by the way, I wonder just how may are coming to Perl 6
without ANY Perl 5 experience?
In every instance I can look up, perldocs puts Perl 6's
documentation to shame.
A simple comparison: which one leaves you knowing how to use
the function and which one leaves you wondering "What the h***???"
$ perldoc -f join
join EXPR,LIST
Joins the separate strings of LIST into a single string with
fields separated by the value of EXPR, and returns that new
string. Example:
my $rec = join(':',
$login,$passwd,$uid,$gid,$gcos,$home,$shell);
Beware that unlike "split", "join" doesn't take a pattern
as its first argument. Compare "split".
https://docs.perl6.org/routine/join#(List)_routine_join
(List) routine join
Defined as:
sub join($separator, *@list --> Str:D)
method join(List:D: $separator --> Str:D)
Treats the elements of the list as strings, interleaves
them with $separator and concatenates everything into a
single string.
Example:
join ', ', <a b c>; # RESULT: «a, b, c»
Note that the method form does not flatten sublists:
say (1, <a b c>).join('|'); # OUTPUT: «1|a b c»
Oh and what the &*@% is a "*@list"? And why does the sub have one
and the method does not? They are suppose to be identical.
-T
