The previous two posts show how to use the ternary operator to assign a value to a variable, but neither mentioned that your lc statement wasn't working as you intend. The lc function will use $_ as an argument if none is provided, but it returns the lowercased string. Since you never assign the return value to anything, it gets ignored (this would have been evident if you used warnings). To fix it, reassign to $_:

print "[$_]\n"; # S
lc( $_ ); # warning: "Useless use of lc in void context"
print "[$_]\n"; # S
$_ = lc( $_ );
print "[$_]\n"; # s
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  $_ eq 's' ? $OPER='s' : $OPER='c';

That's a priority issue. Try:

However, your code has a worse problem: mo strict nor warnings. As a consequence, your usage of lc is useless without you noticing it:

print "Field 1 & 2 Separate or Combined? ";
chomp($_=<STDIN>);
lc($_);
$_ eq 's' ? $OPER='s' : $OPER='c';
print "OPER $OPER\n";
__END__
Field 1 & 2 Separate or Combined? S
OPER c
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Try this fixed version:

use strict;
use warnings;

print "Field 1 & 2 Separate or Combined? ";
chomp ($_ = <STDIN>);
my $OPER = lc eq 's' ? 's' : 'c';
print "OPER $OPER\n";
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$OPER = $_ eq 's' ? 's' : 'c' ;
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This turned into an interesting thread. One quibble: it's called the ternary operator, not the trinary operator.

$OPER = $_ eq 's' ? 's' : 'c';
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Cheers,

JohnGG

There may be a Perl bug here. Because of operator precedence I'd have expected that the trinary experession shouldn't compile. The assignment operator has lower percedence than the trinary operator.

However the immediate answer to your problem is to fix the precedence issue. Here are two fixes:

1/ $test ne 'c' ? ($OPER='s') : ($OPER='c');
2/ $OPER = $test ne 'c' ? 's' : 'c';
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No bug. The statement parses as ($test ne 'c' ? ($OPER = 's') : $OPER) = 'c';

---

Jeff japhy Pinyan, P.L., P.M., P.O.D, X.S.: Perl, regex, and perl hacker
How can we ever be the sold short or the cheated, we who for every service have long ago been overpaid? ~~ Meister Eckhart

which to me looks like:

$OPER = 's' = 'c';
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when $test ne 'c' is true. And that should fail with a "Can't modify constant item in scalar assignment" error. However no errors and no warnings are generated for either condition.

Using ?: to select a variable to be assigned to I'm happy with:

test ? $p1 : $p2 = value;
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but the half assignement variant implied by the precedence (and made explicit by the brackets shown in your reply) is just bizare.

---

DWIM is Perl's answer to Gödel

( $_ eq 's' ) ? ( $OPER = 's' ) : ( $OPER = 'c' ); # still cringeing
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# </cringe>
$OPER = ( $_ eq 's' ) ? 's' : 'c';
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