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lecture02 Manorathan Indran s conflicted copy 2012-02-07 Muk...
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lecture02 Manorathan Indran s conflicted copy 2012...
lecture02_Manorathan_Indran_s_conflicted_copy_2012-02-07_Mukul_Patel_s_conflicted_copy_2012-03-20.ppt-Jie Hu, ECE/NJIT, Spring 2011 ECE252
lecture02 Manorathan Indran s confl...
lecture02_Manorathan_Indran_s_conflicted_copy_2012-02-07_Mukul_Patel_s_conflicted_copy_2012-03-20.ppt-Jie Hu, ECE/NJIT, Spring 2011 ECE252
Page 17
Jie Hu, ECE/NJIT, Spring 2011
ECE252 L02-68000 Arch.17
Endianness: Example
Register value (.L): 0XBDC3A827
0x00..00
0xffffffff
0x00..10
0x00..13
27
A8
C3
BD
0
1
2
3
27
A8
C3
BD
0x00..00
0x00..10
0x00..13
0xffffffff
27
A8
C3
BD
Big Endian
Little Endian
0
31
bit
byte
Page 18
Jie Hu, ECE/NJIT, Spring 2011
ECE252 L02-68000 Arch.18
Number Systems (Review)
Positional number system representation (unsigned):
where
n
is the number of digits,
r
is the radix or base,
and
a
i
are the digits (coefficients),
0 <=
a
i
<
r
Based on the value of the radix
r
, we have
decimal:
r
=10,
a
i
:0,1,2,3,4,5,6.7,8,9
binary:
r
=2,
a
i
:0,1
octal:
r
=8,
a
i
:0,1,2,3,4,5,6.7
hexadecimal:
r
=16,
a
i
:0,1,2,3,4,5,6.7,8,9,A,B,C,D,E,F
In 68000 assembly code, numbers can be either in decimal
(e.g, 91) or hexadecimal (e.g, $5B)
0
1
2
2
2
2
1
...
1
a
r
a
r
a
r
a
r
a
N
n
n
n
n
Page 19
Jie Hu, ECE/NJIT, Spring 2011
ECE252 L02-68000 Arch.19
Base Conversion
From binary/octal/hexadecimal to decimal, use the
representation equation to evaluate
101111
2
=1x2
5
+0x2
4
+1x2
3
+1x2
2
+1x2+1=47
10
1352
8
=1x8
3
+3x8
2
+5x8+2=746
10
2EA
16
=2x16
2
+14x16+10=746
10
Convert decimal to binary
105
10
=(?)
2
105
10
=(1101001)
2
Convert decimal to octal?
105
10
=(?)
8
Convert decimal to hexadecimal?
105
10
=(?)
16
105
2
52
1
2
26
0
2
13
0
2
6
1
2
3
0
2
1
1
2
0
1
Page 20
Jie Hu, ECE/NJIT, Spring 2011
ECE252 L02-68000 Arch.20
Base Conversion
Base conversion among binary, octal, and hexadecimal
binary
octal
1100100101
binary
hexadecimal
1100100101
octal
hexadecimal
001 100 100 101
1
4
4
5
1445
8
0011 0010 0101
3
2
5
325
16
1445
8
001 100 100 101
1
4
4
5
0011 0010 0101
3
2
5
325
16
Page 21
Jie Hu, ECE/NJIT, Spring 2011
ECE252 L02-68000 Arch.21
Signed Binary Representations
Signed-Magnitude Representation
Sign-bit (MSB) indicates a positive (Sign:0) or a negative (Sign:1)
number
the Remaining bits represent the (unsigned) value
e.g., (105)
10
=(0 1101001)
2
, (-105)
10
=(1 1101001)
2
what about (0)
10
= (?)
2
?
(+0)
10
= (0 0000000)
2
, (-0)
10
= (1 0000000)
2
1’s-Complement Representation
Positive number same as in signed-magnitude representation
Derive negative number representation by bit-complementing its
positive representation
e.g., (105)
10
=(0 1101001)
2
, (-105)
10
=(1 0010110)
2
What about (0)
10
?
(+0)
10
= (0 0000000)
2
, (-0)
10
= (1 1111111)
2
Page 22
Jie Hu, ECE/NJIT, Spring 2011
ECE252 L02-68000 Arch.22
Signed Binary Representations
2’s-Complement Representation
Positive number same as in signed-magnitude representation
Derive negative number representation by bit-complementing its
positive representation (x) and
adding
1
to its least significant bit,
i.e., 2
k
– x, where k is the number of bits of the representation
e.g., (105)
10
=(0 1101001)
2
,
e.g., (-105)
10
=(1 0010110 + 1)
2
=(1 0010111)
2
What about (0)
10
?
(+0)
10
= (0 0000000)
2
, (-0)
10
= (1 1111111 + 1)
2
= (0 0000000)
2
What is (1 0000000)
2
=(?)
10
(1 0000000)
2
=(-128)
10
Data range: [-2
k-1
, 2
k-1
-1], for k-bit number representation
68000 uses 2’s-complement representation
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