If you right-shift an 8-bit variable 7 times, you get 1 or 0 according
to the most significant bit.
Regards,
Harri
________________________________
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Dipti
Sent: Monday, April 30, 2007 8:13 AM
To: Jay Taneja
Cc: tinyos forum
Subject: Re: [Tinyos-help] wt does this mean: rcopy >> 7
Hi ,
Thnks for the reply Jay, what i really wanted to ask is why is the value
of the variable rCopy being right shifted 7 times., The IntOutput
interface can output an 8 bit integer and when the value of rCopy is
right shifted 7 times it becomes a no. with 9 bits.. so how can we use
IntOutput interface to output the same?
Dipti Jaiswal
----- Original Message -----
From: Jay Taneja <mailto:[EMAIL PROTECTED]>
To: Dipti <mailto:[EMAIL PROTECTED]>
Cc: tinyos forum <mailto:[email protected]>
Sent: 2007 Apr 30 10:09 AM
Subject: Re: [Tinyos-help] wt does this mean: rcopy >> 7
>> is a bitwise shift operator, meaning shift the value of rCopy
to the right by 7 bits. Here's a website that documents the C operator:
http://www.phim.unibe.ch/comp_doc/c_manual/C/CONCEPT/bit_shift.html#shif
t
Cheers.
-jay
On 4/29/07, Dipti < [EMAIL PROTECTED]
<mailto:[EMAIL PROTECTED]> > wrote:
Hi all,
I was going thru the sensetorfm component. I could not
follow a few lines in the code.
Can someone help me with understanding this..
void task outputTask() {
uint16_t rCopy;
atomic {
rCopy = reading;
}
call IntOutput.output(rCopy >> 7); /* what does
rCopy >>7 denote? */
}
Thanks in advance
Dipti Jaiswal.
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