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).
In reply to Re: Difference of localtime
by almut
in thread Difference of localtime
by vinoth.ree
| For: | Use: | ||
| & | & | ||
| < | < | ||
| > | > | ||
| [ | [ | ||
| ] | ] |