Re: conditional print. Is correct to use it?

G'day pvaldes,

In terms of style, I'd say it's perfectly acceptable to embed an expression in a print list. Having said that, if the expression is long or complicated, I'd usually move it out of the print statement.

$fo ||= 'Roses are red ...';
print "My homework: $fo";
[download]

As others have already pointed out, a FALSE value can be "non empty", so || is probably not the best choice.

# (fo will be updated with a real value in this part)

You'll need to tell us what's going on there. My first guess was that you're perhaps doing something like:

$ perl -E '
    my $fo = "";
    update($fo);
    print "|$fo|";

    sub update {
        my ($arg) = @_;
        $arg .= "XXX";
    }
'
||
[download]

when what you really wanted was something like:

$ perl -E '
    my $fo = "";
    update(\$fo);
    print "|$fo|";

    sub update {
        my ($arg) = @_;
        $$arg .= "XXX";
    }
'
|XXX|
[download]

But there would be dozens of reasons why your update could be returning a "forrible" value. :-)

— Ken

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Re^2: conditional print. Is correct to use it? by AnomalousMonk (Archbishop) on Oct 28, 2020 at 18:03 UTC
... a FALSE value can be "non empty", so `\|\|` is probably not the best choice. I don't understand this statement. (Update: hippo has likely shown the way to understanding. :) Is there any false value of `$x`, empty or not, for which the expression `($x \|\| $y)` would not evaluate as the value of `$y`? Likewise, given the statement `$x \|\|= $y;` is there any false initial value of `$x` for which the final value of `$x` would not end up as `$y`? Give a man a fish: `<%-{-{-{-<`	[reply] [d/l] [select]
Re^3: conditional print. Is correct to use it? by hippo (Bishop) on Oct 28, 2020 at 21:19 UTC
I think kcott is saying that you might, in some conditions, wish to retain the non-empty false value in $x rather than discard it in favour of $y. eg. if 0 were a valid input for $x and you had: `$x = 0; $y = 10; $number = $x \|\| $y; print "$number\n";` [download] You would see 10 rather than the valid-but-false 0. In such cases `$number = $x // $y;` might be more appropriate. 🦛	[reply] [d/l] [select]
Re^3: conditional print. Is correct to use it? by kcott (Archbishop) on Oct 29, 2020 at 03:50 UTC
Pretty much what ++hippo said. :-) length will return FALSE for both `''` and `undef`, and TRUE for both `0` and `'0'`. It's first documented as doing that in 5.12.0; however, there was a bug that was fixed in 5.14.0 (perl5140delta: Syntax/Parsing Bugs) so I'd be more comfortable with both `defined` and `length` if using anything earlier than 5.14.0. # 5.14.0 or later $ perl -E 'my ($x, $y) = (0, "fallback"); $x = length $x ? $x : $y; sa +y $x' 0 $ perl -E 'my ($x, $y) = ("0", "fallback"); $x = length $x ? $x : $y; +say $x' 0 $ perl -E 'my ($x, $y) = ("", "fallback"); $x = length $x ? $x : $y; s +ay $x' fallback $ perl -E 'my ($x, $y) = (undef, "fallback"); $x = length $x ? $x : $y +; say $x' fallback # 5.12.0 or earlier $ perl -E 'my ($x, $y) = (0, "fallback"); $x = (defined $x && length $ +x) ? $x : $y; say $x' 0 $ perl -E 'my ($x, $y) = ("0", "fallback"); $x = (defined $x && length + $x) ? $x : $y; say $x' 0 $ perl -E 'my ($x, $y) = ("", "fallback"); $x = (defined $x && length +$x) ? $x : $y; say $x' fallback $ perl -E 'my ($x, $y) = (undef, "fallback"); $x = (defined $x && leng +th $x) ? $x : $y; say $x' fallback # Probably not what was intended $ perl -E 'my ($x, $y) = ("0", "fallback"); $x \|\|= $y; say $x' fallback $ perl -E 'my ($x, $y) = ("", "fallback"); $x //= $y; say "<$x>"' <> [download] — Ken	[reply] [d/l] [select]


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