Re^2: Evaluating the condition ($x ||= $y)

Am I right?

Yes, I think so.
The problem is fixed by changing the troublesome rendition to:

($x) ? ($z  = "true\n") : ($z  = "false\n"); print $z;
[download]

I sure am glad I generally go for the extra bit of typing and use if/else. Bloody hell ... the amount of time I can spend in a ternary op rabbit hole ... :-(
Thanks harangzsolt33, Fletch, hv.

Cheers,
Rob

Comment on Re^2: Evaluating the condition ($x \|\|= $y) Download Code

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Re^3: Evaluating the condition ($x \|\|= $y) by ikegami (Patriarch) on Aug 26, 2024 at 04:14 UTC
`say $x ? "true" : "false";` [download] And it's called the conditional operator. It is merely a ternary operator, just like addition operator is a binary operator. It not even Perl's only ternary operator.	[reply] [d/l]
Re^4: Evaluating the condition ($x \|\|= $y) by Anonymous Monk on Aug 26, 2024 at 09:31 UTC
It not even Perl's only ternary operator. ???	[reply]
Re^5: Evaluating the condition ($x \|\|= $y) by ikegami (Patriarch) on Aug 26, 2024 at 13:42 UTC
While it's Perl only infix ternary operators, Perl has circumfix ternary operators. `dbmopen`, for one. I would argue that the substitution and translate operators have three operands as well (pattern, replacement and flags).	[reply] [d/l]
Re^3: Evaluating the condition ($x \|\|= $y) by tybalt89 (Monsignor) on Aug 26, 2024 at 07:53 UTC
`#!/usr/bin/perl use strict; # https://perlmonks.org/?node_id=11161348 use warnings; my $x = '1'; my $z = "undetermined\n"; ($x) ? print "true\n" : print "false\n"; ($x) ? $z = "true\n" : $z = "false\n"; print $z; # fewer parens :) $x ? $z = "true\n" : ($z = "false\n"); print $z; # my preferred version $z = $x ? "true\n" : "false\n"; print $z;` [download]	[reply] [d/l]