Re^2: use of parentheses around a variable

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Re^3: use of parentheses around a variable by ikegami (Patriarch) on Jun 24, 2008 at 06:20 UTC
First, keep in mind that `(my $var1, $var2, $var3)` is different than `my ($var1, $var2, $var3)` and `(my $var1, my $var2, my $var3)` The first only declares one variable but the other two declare three. `>perl -Mstrict -e"my ($x, $y)" >perl -Mstrict -e"(my $x, my $y)" >perl -Mstrict -e"(my $x, $y)" Global symbol "$y" requires explicit package name at -e line 1. Execution of -e aborted due to compilation errors.` [download] Now on to your question. What good is making a construct you can't access? That would be quite counter-productive. Either I don't know what you mean by "access", or the answer is "Of course you can access a list".	[reply] [d/l] [select]
Re^3: use of parentheses around a variable by massa (Hermit) on Jun 24, 2008 at 10:32 UTC
`(my $var)` is a list with one scalar called `$var`, and introduces `$var` to the current scope context. `my ($v1, $v2)` is a list with two scalars (`$v1, $v2`) and introduces both to the current scope context. `(my $v1, $v2)` is a list with two scalars, but only introduces `$v1` to the current scope context, meaning that if `use strict` is in effect, `$v2` should be already introduced in the context or an error will occur. When there is a list on the left-hand-side of an assignment, it forces list context to the right-hand-side (IOW, the right-hand-side will be interpreted as a list), so, `my ($v) = @_` reads the first element of `@_` and put it in `$v`. When there is a scalar on the left-hand-side of an assignment, it forces scalar context to the right-hand-side (forces it to be interpreted as a scalar, pretty much as if you had put the word `scalar` in front of it), so, `my $v = @_` is the same as `my $v = scalar @_` and, because an array in scalar context is the number of its elements, this puts the number of elements of `@_` in `$v`. Simple, no? :-)	[reply] [d/l] [select]