Re: [PHP-DEV] why does exit() print its argument? From: Zeev Suraski (zeev <email protected>)
Date: 09/29/01

To make it formal, a strong -1 from me and a +1 on introducing a new function.

Zeev

At 21:06 29-09-01, Markus Fischer wrote:
> > Usualy I am very much against breaking backwards compatibility, but in this
> > case I think it's the best thing to do... Because:
> > - It's already documented that way
> > - It's the 'expected' behaviour (from other languages, and from the docs)
> >
> > As Rasmus said: It would be surprising if this broke a lot of (or even any)
> > code.
>
>+1
>
>Hey, my second vote O_o
>
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