From: "Marin David Condic" <dont.bother.mcondic.auntie.spam@[acm.org>
Subject: Re: "sizeof" a record
Date: Mon, 10 Sep 2001 16:23:04 -0400
Date: 2001-09-10T20:23:06+00:00 [thread overview]
Message-ID: <9nj7fa$e53$1@nh.pace.co.uk> (raw)
In-Reply-To: 9nj40t$qbr$1@zeus.orl.lmco.com
You should be able to do Record_Type'Size or Record_Object'Size which
returns the size in bits. You probably want to divide by System.Storage_Unit
to get the size in bytes. (If your machine has System.Storage_Unit = 8)
Read through Annex K of the ARM - you'll find lots of useful attributes
there that are especially handy for low level programming or dealing with
machine dependencies.
MDC
--
Marin David Condic
Senior Software Engineer
Pace Micro Technology Americas www.pacemicro.com
Enabling the digital revolution
e-Mail: marin.condic@pacemicro.com
Web: http://www.mcondic.com/
"mop" <mop65715@pegasus.cc.ucf.edu> wrote in message
news:9nj40t$qbr$1@zeus.orl.lmco.com...
>
> how do i obtain the "sizeof" a record?
>
> I've got 8 bytes of data for one record
> and 517 bytes for another
>
>
>
>
next prev parent reply other threads:[~2001-09-10 20:23 UTC|newest]
Thread overview: 3+ messages / expand[flat|nested] mbox.gz Atom feed top
2001-09-10 19:24 "sizeof" a record mop
2001-09-10 20:23 ` Marin David Condic [this message]
2001-09-10 20:40 ` David C. Hoos
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