Re: why does exit() print its argument?

From: Date: Sat, 29 Sep 2001 21:32:04 +0000
Subject: Re: why does exit() print its argument?
References: 1 2 3  Groups: php.dev 
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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 -- PHP Development Mailing List <http://www.php.net/> To unsubscribe, e-mail: php-dev-unsubscribe@lists.php.net For additional commands, e-mail: php-dev-help@lists.php.net To contact the list administrators, e-mail: php-list-admin@lists.php.net


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