When $i ne 'one', the constant:  'nothing' is not assigned to anything, thus in a void context.

Did you want:

$hash->{'name'} = ($i eq 'one') ? "hello"."world" : 'nothing';
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Yes , you are right, I want what you just mentioned.
But my question is where does $i ne 'one' is happening in my code ? $i is assgined to 'one' and not getting changed.

But my question is where does $i ne 'one' is happening in my code ? $i is assgined to 'one' and not getting changed.

The point is that the compiler knows at compile-time that 'nothing' is in a void context; so it warns you.

You can choose to ignore that warning; or you could even disable it; but it is surely better to simply correct the code.

---

With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.

"Science is about questioning the status quo. Questioning authority".

In the absence of evidence, opinion is indistinguishable from prejudice.

where does $i ne 'one' is happening in my code ?

It isn’t! To see this, print out the values of $hash->{name} and $i:

#! perl
use strict;
use warnings;

my $hash = {};
my $i    = 'one';

print "\nOriginal code:\n\n";
($i eq 'one') ? $hash->{'name'} = "hello"."world" : 'nothing';
printf "name: %s, i: %s\n", $hash->{name}, $i;

print "\nBrowserUk's fix:\n\n";
$hash->{name} = ($i eq 'one') ? "hello" . "world" : 'nothing';
printf "name: %s, i: %s\n", $hash->{name}, $i;
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Output:

22:46 >perl 801_SoPW.pl
Useless use of a constant ("nothing") in void context at 801_SoPW.pl l
+ine 21.

Original code:

name: helloworld, i: one

BrowserUk's fix:

name: helloworld, i: one

22:46 >
[download]

As you can see, the warning is printed before the script runs, because it’s a compile-time warning, as BrowserUk says. But in both cases, the result of the assignment is that $hash->{name} contains helloworld, showing that $i eq 'one' is true. What makes you think that $i becomes not equal to 'one'?

For the ternary operator, see perlop#Conditional-Operator.

Hope that helps,

Athanasius <°(((>< contra mundum

Iustus alius egestas vitae, eros Piratica,

It sounds almost like you expect c<$i> to be updated - which won't happen; Are you are confusing a test with an assignment - the eq operator tests for string equality between the LHS and RHS, it is most definitely not an assignment operator.
A user level that continues to overstate my experience :-))

if ($expression) {
  $var = "foo";
} else {
  $var = "bar";
}

# WRONG
$expression ? $var = "foo" : $var = "bar";

# correct
$var = $expression ? "foo" : "bar";
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This is bad information.

The expression

$expression ? $var = "foo" : $var = "bar";
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is only "# WRONG" because of operator precedence (see perlop: Operator Precedence and Associativity). Here's how it is parsed:

$ perl -MO=Deparse,-p -e '$expression ? $var = "foo" : $var = "bar";'
(($expression ? ($var = 'foo') : $var) = 'bar');
-e syntax OK
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This is explained in perlop: Conditional Operator. Correcting the precedence with parentheses:

$ perl -MO=Deparse,-p -e '$expression ? ($var = "foo") : ($var = "bar"
+);'
($expression ? ($var = 'foo') : ($var = 'bar'));
-e syntax OK
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I do not dispute that '$var = $expression ? "foo" : "bar";' is a better way to write '$expression ? ($var = "foo") : ($var = "bar");'. In fact, the documentation also recommends this.

Once corrected as indicated, here's how your "# WRONG" code behaves:

#!/usr/bin/env perl -l

use strict;
use warnings;

for my $expression (0, 1) {
    my $var = '';
    $expression ? ($var = 'foo') : ($var = 'bar');
    print "\$expression = '$expression'; \$var = '$var'";
}
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Output:

$expression = '0'; $var = 'bar'
$expression = '1'; $var = 'foo'
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-- Ken

But what does the syntactically correct statement
    $expression ? $var = "foo" : $var = "bar";
actually do?

>perl -wMstrict -le
"for my $expression (0, 1) {
   my $var = '';
   $expression ? $var = 'foo' : $var = 'bar';
   print qq{if \$expression is $expression, \$var is '$var'}
   }
"
if $expression is 0, $var is 'bar'
if $expression is 1, $var is 'bar'
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It seems the statement is just an expensive equivalent to the statement
    $var = 'bar';
The reason why is left as a exercise for simonz (after the recommended perusal of perlop :)

If what you really want is:

• Assign "hello"."world" to the hash when ($i eq 'one')
• Do nothing when ($i ne 'one')

Then one way to achieve this is:

($i eq 'one') ? $hash->{'name'} = "hello"."world" : 0;
[download]

Executing "0" does not give a warning, unlike executing 'nothing'.

Though that would be a hell of a lot more readable as:

$hash->{'name'} = "hello"."world"
   if $i eq 'one';
[download]

use Moops; class Cow :rw { has name => (default => 'Ermintrude') }; say Cow->new->name