Subject:
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Re: Somewhat C challenged.
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Newsgroups:
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lugnet.robotics.handyboard
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Date:
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Thu, 10 Apr 1997 02:48:28 GMT
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Original-From:
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Prabal Dutta <(prabal@dmsc.)AvoidSpam(com)>
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Reply-To:
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prabal@dmsc.+AvoidSpam+com
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Viewed:
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1423 times
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Victor,
The peek(0x1000) statement returns the byte at memory location 1000
(hex). In C, the '&' operator is a bitwise AND. So, what you are doing
is ANDing 4 and the contents of 0x1000. What's really going on is you
are interested in only bit 3 (out of bits 0 through 7), and the
statement is masking the other bits:
mem(0x1000) = xxxx xxxx
4 = 0000 0100
-------------
0000 0x00
That way you only see the value of bit 3.
- Prabal
victor w clark wrote:
>
> Please excuse my bitwise ignorance. Could someone explain this statement
> to me.
> if (4 & peek(0x1000))
> what does peek(0x1000) return?
> what is the 4 for ?
> and what does the "&" compare?
>
> thanks
> victor
--
Prabal Dutta Tel: (614) 299-2566
Digital Micro Systems, Corp. FAX: (614) 299-2939
PO Box 3579 e-mail: prabal@dmsc.com
Columbus OH 43210-0579 WWW: http://www.dmsc.com
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Message is in Reply To:
 | | Somewhat C challenged.
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| Please excuse my bitwise ignorance. Could someone explain this statement to me. if (4 & peek(0x1000)) what does peek(0x1000) return? what is the 4 for ? and what does the "&" compare? thanks victor (27 years ago, 9-Apr-97, to lugnet.robotics.handyboard)
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