set 0xDEADBEEF, %o1
set 0x13579246, %o2
xor %o1, %o2, %o1
What will be in register o1?
set 0xDEADBEEF, %o1
set 0x13579246, %o2
and %o1, %o2, %o1
What will be in register o1?
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If you have SPARC at your disposal, why can't you just try it? – mvp Apr 25 '13 at 08:12
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I am sorry. I understand how to do it. My method of doing it, for example for XOR, convert both numbers to binary, then use the logic if both are 0 or if both are 1, put 0 and if either is different put 1 and create a new binary number, then convert it back to Hex But the problem with that is, I have to do it on a timed test and I do not think I have time to do all that and other questions. So I am looking for a faster way to do this – user2318744 Apr 25 '13 at 08:31
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Just start Linux `gnome-calculator` or Windows `calc` in programmers mode and use `XOR` or `AND` buttons (I am sure Mac has something similar as well). – mvp Apr 25 '13 at 08:51
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Can't use a calculator / computer on the test xDD – user2318744 Apr 25 '13 at 09:05
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1Then just work on it one hex digit at a time - I guess that's what you have been doing anyway. For example, `D xor 1` = `C`, `E xor 3` = `D`, ... – mvp Apr 25 '13 at 09:09
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You have time... first off practice converting from a hex digit directly to binary in your head. So you don't have to look those up at the very least and I do mean going directly from C -> 1101 and so on. – cb88 Apr 30 '13 at 12:41
1 Answers
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CDFA2CA9 and 12059246 respectively, since both are bitwise operations, there is no architecture dependency here. see: http://en.wikibooks.org/wiki/SPARC_Assembly/Arithmetic_Instructions#Logic_Instructions
Regarding the actual calculation:
Note that:
0x0 == 0b0000
0x1 == 0b0001
0x2 == 0b0010
...
0xf == 0b1111
Since in bitwise operations, each digit in any base that is a power of 2 is independnt, it is simple to create a table for any base:
in base 2 (^ means xor, & means and):
& 0 1
0 0 0
1 0 1
^ 0 1
0 0 1
1 1 0
And in base 16 (hexadecimal):
and 0 1 2 3 4 5 6 7 8 9 A B C D E F
0 0 1 2 3 4 5 6 7 8 9 A B C D E F
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1
2 0 0 2 2 0 0 2 2 0 0 2 2 0 0 2 2
3 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3
4 0 0 0 0 4 4 4 4 0 0 0 0 4 4 4 4
5 0 1 0 1 4 5 4 5 0 1 0 1 4 5 4 5
6 0 0 2 2 4 4 6 6 0 0 2 2 4 4 6 6
7 0 1 2 3 4 5 6 7 0 1 2 3 4 5 6 7
8 0 0 0 0 0 0 0 0 8 8 8 8 8 8 8 8
9 0 1 0 1 0 1 0 1 8 9 8 9 8 9 8 9
A 0 0 2 2 0 0 2 2 8 8 A A 8 8 A A
B 0 1 2 3 0 1 2 3 8 9 A B 8 9 A B
C 0 0 0 0 4 4 4 4 8 8 8 8 C C C C
D 0 1 0 1 4 5 4 5 8 9 8 9 C D C D
E 0 0 2 2 4 4 6 6 8 8 A A C C E E
F 0 1 2 3 4 5 6 7 8 9 A B C D E F
xor 0 1 2 3 4 5 6 7 8 9 A B C D E F
0 0 1 2 3 4 5 6 7 8 9 A B C D E F
1 1 0 3 2 5 4 7 6 9 8 B A D C F E
2 2 3 0 1 6 7 4 5 A B 8 9 E F C D
3 3 2 1 0 7 6 5 4 B A 9 8 F E D C
4 4 5 6 7 0 1 2 3 C D E F 8 9 A B
5 5 4 7 6 1 0 3 2 D C F E 9 8 B A
6 6 7 4 5 2 3 0 1 E F C D A B 8 9
7 7 6 5 4 3 2 1 0 F E D C B A 9 8
8 8 9 A B C D E F 0 1 2 3 4 5 6 7
9 9 8 B A D C F E 1 0 3 2 5 4 7 6
A A B 8 9 E F C D 2 3 0 1 6 7 4 5
B B A 9 8 F E D C 3 2 1 0 7 6 5 4
C C D E F 8 9 A B 4 5 6 7 0 1 2 3
D D C F E 9 8 B A 5 4 7 6 1 0 3 2
E E F C D A B 8 9 6 7 4 5 2 3 0 1
F F E D C B A 9 8 7 6 5 4 3 2 1 0
The first method means doing 32 very simple calculations (after a simple conversion to binary and then reconverting to hexadecimal), the second means doing 8 lookups in a table that you may or may not be able to prepare in advance.
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Can you tell me how you got the answer? I am kind of confused on it. – user2318744 Apr 25 '13 at 08:18