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6.1 无符号数和有符号数

Dec 31, 2015

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6.1 无符号数和有符号数. 6.2 数的定点表示和浮点表示. 6.3 定点运算. 6.4 浮点四则运算. 6.5 算术逻辑单元. 6.1 无符号数和有符号数. 寄存器的位数. 反映无符号数的表示范围. 一、无符号数. 8 位 0 ~ 255. 16 位 0 ~ 65535. 0 1011. 小数点的位置. 1 1011. 小数点的位置. 0 1100. 小数点的位置. 1 1100. 小数点的位置. 二、有符号数. - PowerPoint PPT Presentation
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[x] =
x 1 x ≥ 0

:
:

:

x = 0.0000

- 3
6
3
+ 9
6
15
- 12
3
5 ≡ + 7 mod 12
0.1001≡ + 1.0111 mod 2
2n+1 + x 0 x ≥ 2nmod 2n+1
[x] = 27+1 +( 1011000 )
1011000
x = 0.1100000
= 11111 + 1 1010
1
x = 1010
x
x
∴ x = 0.1111
= 1,1110 – 100000
1
[x] [x]
[+ 0] = [ 0]

0x 2n x ≥ 0
( 2n+1 – 1) + x 0 ≥ x 2nmod 2n+1 1
x = 1101
= 11111 1101
= 1.1111 0.1010
[x] =

x = +0.0000
= 1,1110 – 11111


1
00000000
00000001
00000010
<> [y] = 0. y1 y2 yn













± 0 0 0 0 0
+ 1 1 1 1 1
0 0 0 0 0 0
1 1 1 1 1 1
0 0 0 0 0 0
1 0 0 0 0 0
- 1 0 0 0 0 0
- 1 1 1 1 1
- 1 1 1 1 0
- 0 0 0 0 1
± 0 0 0 0 0
+ 0 0 0 0 1
+ 0 0 0 1 0
+ 1 1 1 1 0
+ 1 1 1 1 1





r = 2


–2–( 2m–1)×2–n
2( 2m–1)×( 1 – 2–n)
2–( 2m–1)×2–n
–215 ×( 1 – 2-10)



24 ±3 1
m = 4n = 18

r = 4
r = 8
r



000
19
128
6.13 + 10 1 5 1
19
128



x = – 111010






2( 2m–1)×(1 – 2–n)
2–( 2m–1)×2– n








“ 0 ”
… …
… …
1 8 23 32
1 11 52 64
1 15 64 80








+ 6
0,0000110
+13
0,0001101
+104
0,1101000
+ 52
0,0110100
+26
0,0011010

= +11010


Cy
0
0



= 1 0 . 0 1 1 0
= [A + B]
[A]
[B]
= 1 1 , 0 0 1 0
= [A + B]
[A + B]
[A+B]
A = 15 B = 24 A – B

A – B = + 1110110 = + 118
∴ A – B = – 1001 = –9
9
16
11
16
12
16



00. ×××××
11. ×××××
10. ×××××
01. ×××××
00, ×××××
11, ×××××
10, ×××××
01, ×××××

GS
1 1 0 1
1 1 0 1
0 0 0 0
1 1 0 1


= 0.1A + 0.01[0 • A + 0. 1( A +0.1A)]
= 0.1{A +0.1[ 0 • A+0.1(A + 0.1A)]}
= 2-1{A +2-1[ 0 • A+2-1(A + 2-1(A+0))]}
A + 0
+
= 0
1
1
1
1 1
1 1 1
1 1 1 1
1 0 1 1
1
1


3


1 1

n = 4 4 4






x0 y0




z0 = 0
0 . 0 1 1 1
0
1 0
1 1 0
1 z4
0 1 1 0
1 1 0 1
1 z1
1 z3


x*• y* = 0. 1 0 1 1 0 1 1 0
[x • y] = 1. 1 0 1 1 0 1 1 0




x0 y0 = 1 0 = 1





z–x* 2, y* 2, Cj “1”
z+2x* 2, y* 2, Cj “0”
z+x* 2, y* 2, Cj “0”
+x* +2x* –x* 2
+[x*] +[2x*] +[–x* ] 2
z–x* 2, y* 2, “1” Cj
z+2x* 2, y* 2, “0” Cj
z+x* 2, y* 2, “0” Cj

0 0 . 1 1 1 0 0 1
0
+ 2x*Cj = 0
– x* Cj = 1
+ x* Cj = 0
1 1
0 0 0 1 1 1
1 1 1 . 1 0 0 1 0 0
0 1 1 1
1 1 0 0 1 1 1 0
2
0 1 1 1 0 0 1 1
2
0 0 0 1 1 1 0 0
2
Cj



x*• y* = 0. 1 1 1 0 0 0 0 0 0 1 1 1
[x • y] = 1. 1 1 1 0 0 0 0 0 0 1 1 1
6.22




x0 y0 = 0 1 = 1



[–x]
(1)








Booth
[z0]= 0
[z1]= 2-1{(yn+1–yn)[x]+[z0]} yn+1 = 0
[zn]= 2-1{(y2–y1)[x]+[zn-1]}




1 . 0 1 0 1
0
1
1 1
1 1 1
1 1 1 1
1 1 0 1 0
1
1
1 1 1 0 1
0
1
1 1 1 1 0
1
1
1 1 1 1 1
0
1




0 . 1 0 1 1
0 . 1 1 0 1
0 . 0 1 1 0 1
0 . 0 1 0 0 1
0 . 0 0 1 1 0 1
0 . 0 0 0 1 0 1
0 . 0 0 0 0 1 1 0 1
0 . 0 0 0 0 0 1 1 1
1


– 0. 0 0 0 0 0 1 1 1

2

1

y* = 0.y1y2 yn y



x* y*
x* y*
x
y
x*
y*
+[– y*]


1


0
1
0
0 1
x0 y0 = 1 1 = 0
x
y
[ ]

0

+[– y*]

+[– y*]
0 1
1
0 1 1
0 1 1
0
0 1 1 0
0 1 1 0

Ri0 “0” Ri + y*
2( Ri+y*) – y* = 2Ri + y*



+[– y*]
0
0
1
0 1
0 1 1
0 1 1 0
0 1
0 1 1 0
0 1 1
0
x0 y0 = 1 1 = 0
x*
y*

= 0.1101
x
y
[ ]
∴ = 0.1101

Qn
0 A n
n + 1

0 X n
0 Q n
[x][y]


()“1”
()“0”

()“0”
()“1”



1 . 0 1 0 1
0 . 1 1 0 1
1 . 0 0 1 1
0 . 1 1 0 1
0 . 1 1 0 1
0 . 0 0 0 0

1
“1”
“0”
+[y]
“0”
+[y]
“1”
“1”

1
1 0
1 0 0
1
1
1 0
1 0 0
1 0 0 1


[x] = 00, 01; 00.1101 [y] = 00, 11; 11.0110
1.
= 00, 01
11, 01
11, 10
– 2

+
+
x + y



∴ x + y = (– 0.1110)×210
1 1
01. ×× × 10. ×× ×





= 00, 010
11, 111
100, 001
+1
[Sx] = 00. 110100
[Sy]' = 00. 010110
x = 0.1101× 210 y = 0.1011 × 201
x +y 3 6
∴ Sy 1, jy+1

4.

(2) “1”

= 11, 011
00, 100
11, 111
–1
x = (– 0.101000)×2-101
y = ( 0.111000)×2-100
5
8
7
8
x – y 3 6
∴ Sx 1, jx + 1


5.
2 7 2 n
2127×(–1)
2.
(2)
= Ai Bi + (Ai+Bi)Ci-1
di = Ai Bi
ti = Ai + Bi
Ci = di + tiCi-1
Si = Ai Bi Ci-1+Ai Bi Ci-1+Ai Bi Ci-1+Ai Bi Ci-1
Ci = Ai Bi Ci-1+Ai Bi Ci-1+Ai Bi Ci-1+Ai Bi Ci-1
FAn
FAn-1
FA1
FA0
FAn-2
Cn
Sn
Cn-1
Sn-1
Cn-2
Sn-2


n = 32
8
D8
C-1
1
≥1
C15 C11 C7 C3
C14~C12 C10~C8 C6~C4


D5
T5
D6
T6
D7
T7
D8
T8
C15
C11
C7
C3
C15 C11 C7 C3
C18 ~C16 C14~C12 C10~C8 C6~C4
C31 C27 C23 C19
2.5 ty