Re: Counting the number of items returned by split without using a named array

[Tanalis]

in reply to Counting the number of items returned by split without using a named array

Why not just count the instances of whitespace?

my $str = "this is a test string";
my $cnt = 1;                        # num tokens = whitespace + 1

++$cnt while $str =~ /\s+/g;
print $cnt, "\n";                   # prints 5

[blazar]

The =()= does work with matches:

$ perl -lpe '($_=()=/\s+/g)++'
foo
1
bar baz
2
foo bar baz
3

But I would use split, especially with the smart behaviour provided by the default ' ' argument.

[Hue-Bond]

That fails when input has leading or trailing blanks:

$ perl -lpe '($_=()=/\s+/g)++'   ## leading
 foo bar
3
$ perl -lpe '($_=()=/\s+/g)++'   ## trailing
foo bar 
3
$ perl -lpe '($_=()=/\s+/g)++'   ## both
 foo bar 
4
$ _

It's better to count ocurrences of actual elements (\S+):

$ perl -wle 'print scalar (()=/\S+/g) for "a b c", " a b c", "a b c ",
+ " a b c "'
3
3
3
3
$ _

Anyway I prefer split too, like salva++'s 0e0 solution.

[blazar]

I know. My entire point was not to maintain a counter manually a' la

++$cnt while $str =~ /\s+/g;

Well done to point out about \S anyway, since one often forgets about \S, \D and \W.