Amit,
One thing that may have confused you. Although the c64x supports
"double register operations" [64 bits] for some instructions, the shifts are 40
bit only and that is why the high byte [0x00000083] was treated as a signed
number.
mikedunn
Ajeet K Mall <a...@emuzed.com>
wrote:
pls. try ....... shru a25:a24,diff1a,a25:a24 ; and it
should help... shr shift as signed and hence the prob. Ajeet
-----Original Message----- From: Amit Agarwal [mailto:a...@yahoo.com] Sent: Friday, August 22, 2003 12:54 AM To: c...@yahoogroups.com Subject: [c6x] problem in right shift operationHi
I am programming
C6414. I am getting an error when i am performing a right shift operation
on a register pair.
shr a25:a24,diff1a,a25:a24 ;
i am
getting the correct result, but on this particular case when initally the value in a25:a24 is a25 00000083h a24 2e7e6a8ch and
diff1a is 19h
i am getting incorrect result.the result after shifting
which i should get is a25 00000000h a24 00004197h
but i am
getting a25 000000ff a24 ffffc197hcan anyone please tell me
why i am getting this.
Thanks Amit Agarwal _______________________________________________________________________ Note: If you do a simple "reply" with your email client, only the author of this message will receive your answer. You need to do a "reply all" if you want your answer to be distributed to the entire group.
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