Re: Idiomatic Split to Scalar Conversion

Perlfunc on split says that "If not in list context ... splits into the @_ array". So I guess that the warning is valid.

If you are just counting you could do: my $foo = $bar =~ tr/,/,/ + 1; To stick with split you could do this:

    my $foo = my @a = split /,/, $bar;


    # Or this
    $foo++ for split /,/, $bar;


    # But why doesn't this work?
    my $foo = () = split /,/, $bar;
[download]

--
John.

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Re: Re: Idiomatic Split to Scalar Conversion by sfink (Deacon) on Apr 30, 2002 at 00:11 UTC
`> # But why doesn't this work? > my $foo = () = split /,/, $bar;` [download] Because split is DWIM-happy (or DWYTIM -- Do What You Think I Mean), and it's told the length of the array it is assigning into so it can stop looking after it fills your array. `my $foo = () = split(/,/, $bar, -1)` almost works, but it doesn't strip trailing undefs like it would in list context without the third param (try it with `"a,b,,"`). If you don't mind wiping out `@_`, then just be explicit about it: `my $foo = @_ = split /,/, $bar`	[reply] [d/l] [select]
Re^3: Idiomatic Split to Scalar Conversion by particle (Vicar) on Apr 30, 2002 at 04:29 UTC
interestingly enough, `my $foo = @{[]} = split( /,/, $bar );` [download] seems to work ( and this one's strict- and warnings-happy :-) ~Particle accelerates	[reply] [d/l]