overriding time() if module exists

sfink has asked for the wisdom of the Perl Monks concerning the following question:

I am trying to override the builtin time() function with Time::HiRes::time() if Time::HiRes is available. It seems to be more difficult than I expected, perhaps because it is defined as an XS routine? Or perhaps because it is doing special magic through the perl API to override time() itself?

This works:

*gettime = do { eval "use Time::HiRes"; $@ } ? sub { time } : sub { Ti
+me::HiRes::time() };
print gettime() . "\n";
[download]

But I have to use gettime() instead of time, and I have to use the parentheses. Using a noarg prototype doesn't help. I have tried various combinations of

eval "use Time::HiRes qw(time)"
*CORE::time = ...
*time = ...
sub time () { ... }
BEGIN { eval { use Time::HiRes qw(time) } }

but none of them work.

Is there some way to change the behavior but not the usage of the time() built-in iff the module exists?

Comment on overriding time() if module exists Select or Download Code

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Re: overriding time() if module exists by sgifford (Prior) on Nov 08, 2003 at 04:45 UTC
This works for me: `BEGIN { eval "use Time::HiRes qw(time);"; }` [download]	[reply] [d/l]
Re: Re: overriding time() if module exists by sfink (Deacon) on Nov 08, 2003 at 19:56 UTC
Umm... okay, I have no idea how I missed that. Thank you. That works perfectly. I'm going to go hide my face in a closet now.	[reply]
Re: Re: Re: overriding time() if module exists by ysth (Canon) on Nov 09, 2003 at 03:57 UTC
No need for the closet. The fact that you need to override built-ins at compile time (or at least before the compilation of the code for which you want the override to work) is not well documented.	[reply]
Re: overriding time() if module exists by Anonymous Monk on Nov 08, 2003 at 04:33 UTC
You don't need use, you don't need those "subs"s. `*gettime = eval { require Time::HiRes; 1 } ? \&Time::HiRes::time : \&t +ime;` [download]	[reply] [d/l]