Re: fork() under MS/DOS?

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#2422

From: Eric Hughes
Date: Fri, 28 Sep 1990 18:41

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In article <69400001@suna2> jaswal@suna2.cs.uiuc.edu writes:
>Windows does non-preemptive scheduling, meaning that
>concurrent tasks must voluntarily give up control of the processor
>before other tasks get a time slice.  (I'm not positive of this.)
>This may have changed with Windows v3.0 or when running on a 286 
>or 386.

From _Microsoft Windows: a guide to Programming_, page 1-16 (this
book is from Microsoft Press and include version 3):

"1.6 Tips for Writing Windows Applications 

...  

"Do not take exclusive control of the CPU--it is a shared resource.
Although Windows is a multitasking system, it is non-preemptive. This
means it cannot take control back from an application until the
application releases it.  A cooperative application carefully manages
access to the CPU and gives other applications ample opportunity to
execute."

"Cooperative application" is _such_ newspeak.

Eric Hughes
hughes@ocf.berkeley.edu

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From: hughes@volcano.Berkeley.EDU (Eric Hughes)
Newsgroups: comp.os.msdos.programmer
Subject: Re: fork() under MS/DOS?
Message-ID: <HUGHES.90Sep28114127@volcano.Berkeley.EDU>
Date: 28 Sep 90 18:41:27 GMT
References: <10718@life.ai.mit.edu> <69400001@suna2>
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In-Reply-To: jaswal@suna2.cs.uiuc.edu's message of 18 Sep 90 01:49:00 GMT

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