Paul Johnson <p...@pjcj.net> writes: > On Tue, Sep 09, 2014 at 03:12:00PM -0700, Jim Gibson wrote: >> >> On Sep 9, 2014, at 2:57 PM, Shawn H Corey wrote: >> >> > On Tue, 09 Sep 2014 23:09:52 +0200 >> > lee <l...@yun.yagibdah.de> wrote: >> > >> >> my $i = 1; >> >> my $f = 2.5; >> >> my $s = 'string'; >> >> my $list = (1, 2, 3); >> > >> > No, the count of items in the list gets stored in $list: $list == 3 >> >> Unless the thing on the right-hand-side of the assignment is a 'list' and >> not an 'array'. This is the one place I can think of where the distinction >> between 'list' and 'array' actually makes a difference. >> >> Compare this: >> >> my $n = ( 4, 5, 6 ); >> >> with this: >> >> my @a = ( 4, 5, 6 ); >> my $n = @a; >> >> (Don't try this with the list ( 1, 2, 3 ), either!) > > > There's a little bit of confusion here. When you write >> >> my $list = (1, 2, 3); > there isn't actually a list anywhere in that statement, appearances to > the contrary notwithstanding. > > To understand what is happening, notice first that the assignment is to > a scalar (confusingly named $list in this case). That means that the > assignment is in scalar context and so the expression (1, 2, 3) is > evaluated in scalar context. > > In scalar context the parentheses () are used for controlling > precedence. The values 1, 2 and 3 are just scalar values. And the > commas are comma operators in scalar context. > > In scalar context the comma operator evaluates its left-hand side, > throws it away and returns the right-hand side.
What is the useful use for this operator? > This means that the value of (1, 2, 3) in scalar context is 3, and this > is what gets assigned to $list. > > What is not happening at all is the creation of a list of numbers and a > calculation of its length. > > See also perldoc -q 'difference between a list and an array' How do you convert an array into a list? -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
