(Unknown Poster) wrote in
news: om:
> I've read this section (page 253 of the 3rd edition of Programming
> Perl) enough to confuse myself quite nicely.
>
> One sentence reads, "If the next thing after the arrow is a bracket or
> a brace, the left operand is treated as a reference to an array or
> hash, respectively ..."
>
> I just don't see how this code, and its explanation, square with that
> quote:
>
> print $array[3]->{"English"}->[0];
>
> " ... the fourth element of @array is intended to be a hash reference
> ..."
>
> Isn't it the fourth element of the array referenced by (the poorly
> named) $array ?
Nope. "$array" all by itself refers to a scalar variable that has nothing
to do with @array. "$array[3]", OTOH, means "the fourth element of
@array". It's followed by an arrow, and then an opening brace so, voila!
it's treated as a reference to a hash. This means that
"$array[3]->{"English"}" will be whatever value the hash being referenced
has for the key "English". And that in turn is followed by an arrow
followed by a bracket, so the hash value is now treated as a reference to
an array (which almost certainly isn't @array), and the whole expression
retrieves the value of the first element of that array.
> When there is more than one arrow operator in an expression, is it
> true that only the "initial funny character" is omitted?
>
> When the arrow operator is removed when it would be "between brackets
> or braces, or between a closing bracket or brace and a parenthesis for
> an indirect function call", is the meaning of the expression changed
> in any way?
No.
> Anyone know what is meant by "ordinary arrays"?
The kind of array that you name "@array" or "@emails" or what have you.
> I'll admit to be so baffled by the final "listref" expressions in this
> section that I can't think of a good question.
perldoc perlreftut might help unconfuse you.
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