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

> 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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