use module if condition is met

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Re: use module if condition is met by tcf22 (Priest) on Jan 23, 2004 at 17:06 UTC
Perhaps something like this: `if($type eq 'x'){ eval "use Some::Module"; } eval { mysubtocheck(); }; if($@){ #Sub ran }else{ #Sub not present or died }` [download] - Tom	[reply] [d/l]
Re: use module if condition is met by broquaint (Abbot) on Jan 23, 2004 at 17:08 UTC
For conditional a `use` statement, see the `if` module. But if you're happy to include your modules at runtime then a simple `require` will be be. Finally, to check for the existence of a subroutine you can simply use `defined` e.g `defined &mysubtocheck && mysubtocheck();` [download] Although make sure you use parens when calling a dynamically defined subroutine otherwise you'll get a compile-time error under `strict 'subs'`. HTH `_________ broquaint`	[reply] [d/l]
Re: use module if condition is met by cfreak (Chaplain) on Jan 23, 2004 at 17:12 UTC
`use` requires code in at compile time so what you have really won't work. Your could use `require` which will do it at run-time but is slower. The best approach I've seen (believe this is in the Camel Book) is to use a `BEGIN` block in combination with `eval`. However that is going to only test if the module is installed, I don't believe variables such as your `$type` would be initialized yet (I could be wrong on that one) Then of course you have the problem with your subs. You could create an `AUTOLOAD` in your modules that would capture any calls to subs that didn't exist and return some kind of flag to that affect. Honestly though I think all that is really more work that you want. I think `require`ing the modules is going to be slower than just `use`ing all the ones you may need even if you don't make calls to all of them. Then you don't have to worry about undefined subs either. Lobster Aliens Are attacking the world!	[reply] [d/l] [select]
Re: Re: use module if condition is met by perrin (Chancellor) on Jan 23, 2004 at 17:46 UTC
Not that it really matters, but `require` is not slower than `use`. If anything, it would be faster, since it does less.	[reply] [d/l] [select]
Re: use module if condition is met by Abigail-II (Bishop) on Jan 23, 2004 at 17:16 UTC
`use if $type eq 'x', Some::Module;` [download] Of course, that requires the value of `$type` to be known at compile time. Otherwise, do something like: `if ($type eq 'x') { require Some::Module; Some::Module -> import if Some::Module -> can ("import"); }` [download] Abigail	[reply] [d/l] [select]
Re: use module if condition is met by ysth (Canon) on Jan 23, 2004 at 17:34 UTC
If $type is known when your script is compiling (e.g. based on operating system or information in @ARGV), use the if module for this: `use if $type eq 'x', "Some::Module";` [download] (Your try at `if ($type eq 'x') { use Some::Module }` doesn't work because the use will be performed at compilation time, before the check of $type even happens.) Otherwise, use some form like eval "use $module" or eval { require $module; $module->import; } (but not eval { use $module }, because there the use is again performed at compilation time). Having the use happen at run-time is not ideal, because there can be differences in how your code will compile; e.g.: `$perl -we'eval "use Errno qw/EINVAL/"; print EINVAL' Name "main::EINVAL" used only once: possible typo at -e line 1. print() on unopened filehandle EINVAL at -e line 1. $perl -we'use Errno qw/EINVAL/; print EINVAL' 22` [download] To detect if a subroutine exists, use `exists(&subname)`. This will not detect subs that would be AUTOLOADed, unless they have a forward declaration (e.g. `sub Foo;`). `defined(&subname)` is slightly different and probably not what you want; see perldoc defined and exists.	[reply] [d/l] [select]
Re: Re: use module if condition is met by Grygonos (Chaplain) on Jan 24, 2004 at 02:41 UTC
Thanks monks, the `if` module worked perfectly. I simply passed my client name as `$ARGV[0]` and it works. Grygonos	[reply] [d/l] [select]


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