Re: why does exit() print its argument?

Is there any good reason not to simply add a new function? It's not as if it ever behaved that way. This magic tends to later fire back at us. Zeev At 03:17 29-09-01, Rasmus Lerdorf wrote:

How about if we overload it a bit. I think anybody who does exit(1) is expecting 1 to be set as the return status whereas someone who does exit('something bad happened') is expecting the string to be shown a-la die(). So let's just check the arg and do the appropriate thing. I would be very surprised if that broke anything. -Rasmus On Sat, 29 Sep 2001, Zeev Suraski wrote: At 03:10 29-09-01, Jim Winstead wrote:

On Sat, Sep 29, 2001 at 02:12:51AM +0200, Zeev Suraski wrote:

Uhm, I'm not dumb :) Of course it enables you to do that, but

that's the

wrong place to put it. exit() has, since its introduction, been a

complete

equivalent of die(), and had nothing to do with exit statuses. Putting this code there was wrong (even though who knows, maybe I even did

that).

no, sascha did. php3 does the same thing.

Ok, looks like it's old enough (as old as mid 1999). We should probably leave it at that and introduce a new shell_exit() function. Zeev