Download Lecture9 - Crystal

Computer Architecture CSE 3322
Lecture 9
TEST 1 – Tuesday March 3
Lectures 1 - 8, Ch 1,2
Web Site
crystal.uta.edu/~cse3322
Branch Addressing
beq $s1, $s2, Label
if ($s1 = =$s2) go to Label
op
rs
rt
address
4
17
18
address
6 bits
5 bits
5 bits
16 bits
effective 32 bit address = ( PC + 4 ) + ( address * 4 )
Next Instruction
address is the relative number of instructions
address * 4 = address : 00
This is PC-relative addressing
jump
j Label
Addressing in Jumps
go to Label
op
address
2
address
6 bits
26 bits
The complete 32 bit address is :
address
4 bits
26 bits
00
2 bits
Upper 4 bits of the Program Counter, PC
jump uses word addresses
address * 4 = address:00
This is Pseudodirect Addressing.
Note: 256 MB word boundaries – 64M Instructions
• Where we've been:
– Performance (seconds, cycles, instructions)
– Abstractions:
Instruction Set Architecture
Assembly Language and Machine Language
• What's up ahead:
– Implementing the Architecture
Possible Representations of Negative
Numbers
Sign Magnitude:
000 = +0
001 = +1
010 = +2
011 = +3
100 = -0
101 = -1
110 = -2
111 = -3
One's
Complement
000 = +0
001 = +1
010 = +2
011 = +3
100 = -3
101 = -2
110 = -1
111 = -0
Two's
Complement
000 = +0
001 = +1
010 = +2
011 = +3
100 = -4
101 = -3
110 = -2
111 = -1
• Issues: balance, number of zeros, ease of operations
• Which one is best? Why?
Representing Numbers in Binary
Consider n-1 bits n-1, n-2 ..., 3, 2, 1. ( bit n will be sign bit)
n 1
c   ci 2
i 1
i 1
 cn 1 2
n2
 cn  2 2
n 3
 ...  c2 2  c1 2
1
0
Representing Numbers in Binary
Consider n-1 bits n-1, n-2 ..., 3, 2, 1. ( bit n will be sign bit)
n 1
c   ci 2
i 1
i 1
cmax  2
n 1
 cn 1 2
1
n2
 cn  2 2
n 3
 ...  c2 2  c1 2
1
0
For positive numbers cn  0
For negative numbers, use Two’s Complement, defined as
2n  c
Representing Numbers in Binary
For negative numbers, use Two’s Complement, defined as
2 c
n
Two's
Complement
000 = +0
001 = +1
010 = +2
011 = +3
100 = -4
101 = -3
110 = -2
111 = -1
2 c
Binary
8 – 4= 4
8 – 3= 5
8 – 2= 6
8 – 1= 7
100
101
110
111
n
Representing Numbers in Binary
Consider n-1 bits n-1, n-2 ..., 3, 2, 1. ( bit n will be sign bit)
n 1
c   ci 2
i 1
i 1
 cn 1 2
n2
 cn  2 2
n 3
 ...  c2 2  c1 2
1
0
For negative numbers, use Two’s Complement 2  c
n
2  c  (2  1)  1  c
n
n
1000...0
1
0111...1
Representing Numbers in Binary
Consider n-1 bits n-1, n-2 ..., 3, 2, 1. ( bit n will be sign bit)
n 1
c   ci 2
i 1
i 1
 cn 1 2
n2
 cn  2 2
n 3
 ...  c2 2  c1 2
1
0
For negative numbers, use Two’s Complement 2  c
n
2  c  (2  1)  1  c  (111...1)2  (cn1 ...ci ...c1 )2  1
n
n
Representing Numbers in Binary
Consider n-1 bits n-1, n-2 ..., 3, 2, 1. ( bit n will be sign bit)
n 1
c   ci 2
i 1
i 1
 cn 1 2
n2
 cn  2 2
n 3
 ...  c2 2  c1 2
1
0
For negative numbers, use Two’s Complement 2  c
n
2  c  (2  1)  1  c  (111...1)2  (cn1 ...ci ...c1 )2  1
n
n
if ci  0 , 1  ci  1
Representing Numbers in Binary
Consider n-1 bits n-1, n-2 ..., 3, 2, 1. ( bit n will be sign bit)
n 1
c   ci 2
i 1
i 1
 cn 1 2
n2
 cn  2 2
n 3
 ...  c2 2  c1 2
1
0
For negative numbers, use Two’s Complement 2  c
n
2  c  (2  1)  1  c  (111...1) 2  (cn1...ci ...c1 ) 2  1
n
n
if ci  0 , 1  ci  1
if ci  1 , 1  ci  0 so, 1  ci  ci
Representing Numbers in Binary
Consider n-1 bits n-1, n-2 ..., 3, 2, 1. ( bit n will be sign bit)
n 1
c   ci 2
i 1
i 1
 cn 1 2
n2
 cn  2 2
n 3
 ...  c2 2  c1 2
1
0
For negative numbers, use Two’s Complement 2  c
n
2  c  (2  1)  1  c  (111...1) 2  (cn1...ci ...c1 ) 2  1
n
n
if ci  0 , 1  ci  1
if ci  1 , 1  ci  0 so, 1  ci  ci
Two’s Complement can be formed by Complementing
each bit and adding 1.
Representing Numbers in Binary
Consider n-1 bits n-1, n-2 ..., 3, 2, 1. ( bit n will be sign bit)
n 1
c   ci 2
i 1
i 1
 cn 1 2
n2
 cn  2 2
n 3
 ...  c2 2  c1 2
1
0
For negative numbers, use Two’s Complement 2  c
n
What is the Two’s Complement of a Negative Number?
Representing Numbers in Binary
Consider n-1 bits n-1, n-2 ..., 3, 2, 1. ( bit n will be sign bit)
n 1
c   ci 2
i 1
i 1
 cn 1 2
n2
 cn  2 2
n 3
 ...  c2 2  c1 2
1
0
For negative numbers, use Two’s Complement 2  c
n
What is the Two’s Complement of a Negative Number?
2 n – 2 n – c = 2n – 2 n + c = c
Since cmax  2 n 1  1
Two’s Complement Addition
For x > 0 and y < 0
x  y  x  2n  y  2n  x  y
2 c
n
Two’s Complement Addition
2 c
n
For x > 0 and y < 0
n
n
xy  x 2  y  2  x  y
if x  y then 2n requires cn 1 so, x  y is left
Two’s Complement Addition
2 c
n
For x > 0 and y < 0
x  y  x  2n  y  2n  x  y
if x  y then 2n requires c n 1 so, x  y is left
if x  y then 2n  ( y  x ) the two' s complement of y  x
Two’s Complement Addition
2 c
n
For x > 0 and y < 0
x  y  x  2n  y  2n  x  y
if x  y then 2n requires c n 1 so, x  y is left
if x  y then 2n  ( y  x ) the two' s complement of y  x
For x < 0 and y < 0
x  y  2  x  2  y  2  [2  ( x  y )]
n
n
n
n
2 n requires c n 1 is lost and the two' s complement
of x  y remains
Normal addition works!
For n = 6 : weights : 32, 16, 8, 4, 2, 1
- 31 < numbers < +31
25 = 011001
100110 complement
For n = 6 : weights : 32, 16, 8, 4, 2, 1
- 31 < numbers < +31
25 = 011001
100110 complement
-25 = 100111
For n = 6 : weights : 32, 16, 8, 4, 2, 1
- 31 < numbers < +31
25 = 011001
100110 complement
-25 = 100111 = 2 6-25 = 64 – 25 = 39
For n = 6 : weights : 32, 16, 8, 4, 2, 1
- 31 < numbers < +31
25 = 011001
100110 complement
-25 = 100111 = 2 6-25 = 64 – 25 = 39
15 = 001111
110000 complement
-15 = 110001
For n = 6 : weights : 32, 16, 8, 4, 2, 1
- 31 < numbers < +31
25 = 011001
100110 complement
-25 = 100111 = 2 6-25 = 64 – 25 = 39
15 = 001111
110000 complement
-15 = 110001 = 64 – 15 = 49
For n = 6 : weights 32, 16, 8, 4, 2, 1
- 31 < numbers < +31
25 = 011001
100110 complement
-25 = 100111 = 2 6-25 = 64 – 25 = 39
15 = 001111
110000 complement
-15 = 110001 = 64 – 15 = 49
-25 + 15
100111
001111
110110
For n = 6 : weights 32, 16, 8, 4, 2, 1
- 31 < numbers < +31
25 = 011001
100110 complement
-25 = 100111 = 2 6-25 = 64 – 25 = 39
15 = 001111
110000 complement
-15 = 110001 = 64 – 15 = 49
-25 + 15
100111
001111
110110
001010
-10
For n = 6 : weights 32, 16, 8, 4, 2, 1
- 31 < numbers < +31
25 = 011001
100110 complement
-25 = 100111 = 2 6-25 = 64 – 25 = 39
15 = 001111
110000 complement
-15 = 110001 = 64 – 15 = 49
-25 + 15
100111
001111
110110
001010
-10
39
15
54
64 - 54
For n = 6 : weights 32, 16, 8, 4, 2, 1
- 31 < numbers < +31
25 = 011001
100110 complement
-25 = 100111 = 2 6-25 = 64 – 25 = 39
15 = 001111
110000 complement
-15 = 110001 = 64 – 15 = 49
-25 + 15
100111
001111
110110
001010
-10
25 +(-15)
011001
110001
001010
+10
For n = 6 : weights 32, 16, 8, 4, 2, 1
- 31 < numbers < +31
25 = 011001
100110 complement
-25 = 100111 = 2 6-25 = 64 – 25 = 39
15 = 001111
110000 complement
-15 = 110001 = 64 – 15 = 49
-25 + 15
100111
001111
110110
001010
-10
25 +(-15)
011001
110001
001010
+10
(-15)+(-14)
110001
110010
100011
For n = 6 : weights 32, 16, 8, 4, 2, 1
- 31 < numbers < +31
25 = 011001
100110 complement
-25 = 100111 = 2 6-25 = 64 – 25 = 39
15 = 001111
110000 complement
-15 = 110001 = 64 – 15 = 49
-25 + 15
100111
001111
110110
001010
-10
25 +(-15)
011001
110001
001010
+10
(-15)+(-14)
110001
110010
100011
011101
- 29
For n = 6 : weights 32, 16, 8, 4, 2, 1
- 31 < numbers < +31
25 = 011001
100110 complement
-25 = 100111 = 2 6-25 = 64 – 25 = 39
15 = 001111
110000 complement
-15 = 110001 = 64 – 15 = 49
-25 + 15
100111
001111
110110
001010
-10
25 +(-15)
011001
110001
001010
+10
(-15)+(-14)
110001
110010
100011
011101
- 29
(-25)+(-15)
100111
110001
011000
+ 24???