Hi lee, On Sun, 24 Jan 2016 13:11:37 +0100 lee <l...@yagibdah.de> wrote:
> 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? > Well, I believe its use was originally inherited from https://en.wikipedia.org/wiki/C_%28programming_language%29 where one can do something like: x = (y++, y+2); In Perl 5 though it is preferable to use do { ... } instead: $x = do { $y++; $y+2; }; See http://perldoc.perl.org/functions/do.html . GCC and compatible compilers have a similar feature to Perl 5's do {...} called statement expressions: https://gcc.gnu.org/onlinedocs/gcc/Statement-Exprs.html . > > 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? > You just put it in list context. For example (untested): sub f { print (@_); } my @input = (3, 44, 505, 6.6); f(@input); my @other_list = (5,@input,24); === Regards, Shlomi Fish -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
