@byjr-k
2016-05-31T09:13:09.000000Z
字数 2644
阅读 1840
EEE310 往年试题
大学学习
2014-15
Q1
a) Branch instruction 0x7B001222, located at memory address 0x00009000. What is its mnemonic code?
0x7B001222 = 0111 101 1 000000000001001000100010
offset * 4 + 4 + current
00000000 00000000 01001000 10001100
+ 00000000 00000000 10010000 00000000
00000000 00000000 11011000 10001100
D88C
BLVC #0x0000D88C
b) 0x2297C0AB, what is its mnemonic code?
0x2297C0AB = 0010 00 1 0100 1 0111 1100 0000 10101011
cond op S Rn Rd #rot immidiate
ADDCSS r12, r7, #0xAB
c) LDRNE r11, [r6], #0xC5, what is the machine code?
0001 01 0 0 1 0 0 1 0110 1011 000011000101
cond # P U B W L Rn Rd offset
1496B0C5
如果没有任何对于 flag 的操作，则 cond 是 always 1110
Q2
a) What's the meaning of A, B, Carry-in and Carry-out?
A: running total
B:
Carry-in: the carry value from last calculation
Carry-out: the carry value for the next calculation
b) 解释 Table 1 第三行
carry-out = 1 -> x4
LSL = #2N+1 -> x2
A - B -> x4 - x2 = x2
c) 解释 Table 1 第四行和倒数第一行
carry-in = 0, Mul = 3 -> x3 + 0 = x3
carry-out = 1 -> x4
LSL = #2N -> x1
A - B -> x4 - x1 = x3
carry-in = 1, Mul = 3 -> x3 + 1 = x4
carry-out = 1 -> x4
LSL = #2N -> x1
A + B -> x4 + 0 = x4
d) 解释 Table 1 倒数第二行
carrr-in = 1, Mul = 2 -> x2 + 1 = x3
carry-out = 1 -> x4
LSL = #2N -> x1
A - B -> x4 - x1 = x3
e) 解释两个表的区别
| Carry-in | Multiplier | Table 1 | Table 2 |
|---|---|---|---|
| 0 | 2 | LSL=2N+1(x2), out=1(x4), A-B -> x2 | LSL=2N+1(x2), out=0, A+B -> x2 |
f)
0x5D = 01011101 multiplicand
0x4E = 01001110 multiplier
Table 2:
N in mul LSL ALU out running total
0 0 10 #1 A+B 0 00000000 00000000 00000000 10111010
1 0 11 #2 A-B 1 11111111 11111111 11111110 01000110
2 1 00 #4 A+B 0 00000000 00000000 00000101 00010110
3 0 01 #6 A+B 0 00000000 00000000 00011100 01010110
00011100 01010110 = 1C56
Q3
a) 4-bit ripple carry adder
b) 32-bit carry look-ahead adder
c) 32-bit carry select adder
Q4
b)
ADD r4, r3, r3, LSL#5
c)
RSB r3, r3, r3, LSL#4
MOV r4, r3, ASR#5
Appendix of 2014-15
2013-14
Q1
a)
| Multiplier | Multiplicand | total |
|---|---|---|
| - | - | 0 |
| 01011100 | 00000000 10110111 | 00000000 00000000 |
| 00101110 | 00000001 01101110 | 00000000 00000000 |
| 00010111 | 00000010 11011100 | 00000010 11011100 |
| 00001011 | 00000101 10111000 | 00001000 10010100 |
| 00000101 | 00001011 01110000 | 00010100 00000100 |
| 00000010 | 00010110 11100000 | 00010100 00000100 |
| 00000001 | 00101101 11000000 | 01000001 11000100 |
0xB7 = 10110111 = 183
0x5C = 01011100 = 92
0x41C4 = 01000001 11000100 = 16836
b)
Multiplicand = 0xB7 = 10110111
Multiplier = 0x5C = 01011100
| N | in | Mul | LSL | ALU | out | total |
|---|---|---|---|---|---|---|
| 0 | 0 | 00 | 0 | A+0 | 0 | 00000000 00000000 |
| 1 | 0 | 11 | 2 | A-B | 1 | 11111101 00100100 |
| 2 | 1 | 01 | 5 | A+B | 0 | 00010100 00000100 |
| 3 | 0 | 01 | 6 | A+B | 0 | 01000001 11000100 |
01000001 11000100 = 16384 + 256 + 128 + 64 + 4 = 16836
c)
SUB r1, r1, r1, LSL#4
MOV r2, r1, ASR#5
Q2
a) mnemonic code of 0x25A3A00C
0x25A3A00C = 0010 01 0 1 1 0 1 0 0011 1010 000000001100
cond # P U B W L Rn Rd offset
STRCS r10, [r3, #-12]!
b) decrease clock cycle; increase instruction density
LDM 7+2=9
STM 7+1=8
c)
LDMFA r13!, {r2, r5, r7, r14}
1110 100 1 1 0 1 1 1101 0100000010100100
cond 100 P U S W L Rn register list
d)
0x73B0C0DE = 0111 00 1 1101 1 0000 1100 0000 11011110
cond 00 # op S Rn Rd #rot immidiate
MOVVCS r12, r0, #0xDE
Q3
a)
MOV r1, #0x7000
MOV r2, #0x9000
MOV r3, #100
loop LDR r4, [r1], #4
ADD r4, r4, #10
STR r4, [r2], #4
SUBS r3, r3, #1
BNE loop
b) 3 state
LDR 总共 5 cycle
STR 总共 4 cycle
BNE 总共 3 cycle，需要 E 之后才能开始下一个的 F
c) 5 state
Appendix of 2013-14
