Difference of localtime

[vinoth.ree]

my $now = localtime time; print "It is now $now"; my $re= scalar localtime; print "It is now $re"; In the above code any different is there or both are same.

[almut]

Depends on what you mean by "code any different?".

In case you want to know if the internally generated opcodes are the same, you can use perl -MO=Concise ... to show them (see B::Concise).  In this case you'd see that the code is different, i.e. scalar is still there (not optimised away, Update: see ysth's note) in the one case, despite being redundant, and time is being called explicitly in the other case, despite being the default with localtime anyway:

...
2     <;> nextstate(main 1 745882.pl:4) v ->3
6     <2> sassign vKS/2 ->7
4        <1> localtime[t3] sK/1 ->5
3           <0> time[t2] s ->4
5        <0> padsv[$now:1,2] sRM*/LVINTRO ->6
7     <;> nextstate(main 2 745882.pl:5) v ->8
...

vs.

...
2     <;> nextstate(main 1 745882.pl:7) v ->3
5     <2> sassign vKS/2 ->6
-        <1> scalar sK/1 ->4
3           <0> localtime[t2] s ->4
4        <0> padsv[$re:1,2] sRM*/LVINTRO ->5
6     <;> nextstate(main 2 745882.pl:8) v ->7
...

That said, the code will produce the same output... (if you happen to execute both statements immediately one after another, and unless you happen to pass a seconds boundary in between the two calls).

[ysth]

Perl's opcodes are arranged in an execution tree as well as the parse tree. The - to the left indicates that scalar was optimized away and is not in the execution tree. -MO=Concise,-exec shows only the execution tree.

The s to the right of localtime indicates scalar context in both cases.

[ikegami]

Furthermore, scalar is optimized away even when it isn't redundant, since scalar only has a compile-time effect.

>perl -MO=Concise -e"@a = scalar @b"
a  <@> leave[1 ref] vKP/REFC ->(end)
1     <0> enter ->2
2     <;> nextstate(main 1 -e:1) v:{ ->3
9     <2> aassign[t5] vKS ->a
-        <1> ex-list lK ->6
3           <0> pushmark s ->4
-           <1> scalar sK/1 ->-          <--- Note leading dash
5              <1> rv2av[t4] sK/1 ->6    <--- "s" means "scalar"
4                 <#> gv[*b] s ->5            did its job.
-        <1> ex-list lK ->9
6           <0> pushmark s ->7
8           <1> rv2av[t2] lKRM*/1 ->9
7              <#> gv[*a] s ->8
-e syntax OK

It's impossible for scalar to work at run-time, since operands are evaluated before their operator.

[ww]

perl -e "my $now = localtime time; print \"It is now $now \"; my $re= scalar localtime; print \"It is now $re \";" on Windows...

or, for a nix-ish OS...

perl -e 'my $now = localtime time; print "It is now $now"; my $re= scalar localtime; print "It is now $re";'

Go ahead. Try it. In fact, you can often answer your own questions, by such trials.

Or did you mean to ask a question that's not evident, such as "Does using scalar make a difference in the output? (Answer: not here) or "Does changing the variable's name make a difference?" (Answer: $now is certainly easier to understand than $re) or "Will one be quicker than the other?" (Answer: Benchmark it.)

And please, you've been here long enough to know that code -- even snippets like these -- should be wrapped in <c>...</c> tags.

Update: Fixed missing </p> tag; clarified phrasing of first question.

[Anonymous Monk]

Or for both

perl -e warn-localtime(time);warn-localtime

-Tue Feb 24 00:35:27 2009 at -e line 1.
-Tue Feb 24 00:35:27 2009 at -e line 1.

[Lawliet]

$ perl -E 'say localtime time; say scalar localtime'
1946222311091530
Mon Feb 23 22:46:19 2009

$ perl -E'$foo = localtime time; say $foo; $bar = localtime; say $bar'
Tue Feb 24 16:34:31 2009
Tue Feb 24 16:34:31 2009

You be the judge! :) (This time, be the judge with misleading code. Thanks, ysth and Annony.)

I guess I should have suggested for you to just try it yourself to begin with~

Update: Added smiley

[Anonymous Monk]

That is misleading because  my $foo = localtime; is scalar context.