S0 S1 S2 S3 S4 S5 S6 S7

Cours Architecture et technologie des ordinateurs II
Exercices - Semaine IV - Solutions
Question 1:
A = R1
B = R2
C = R3
Résultat: C = A + B
S0
start=0
S7
outport ← C
start≠0
S1
A ← inport
S6
C ← C >> 1
S2
B ← inport
S5
C ← C + B
S3
C ← A + B
S4
C ← C + A
State Graph
S0
S1
S2
S3
S4
S5
S6
S7
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| O
start | E
0 1 | N
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S0 S1 | 0
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S2 S2 | 0
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S3 S3 | 0
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S4 S4 | 0
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S5 S5 | 0
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S6 S6 | 0
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S7 S7 | 0
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S0 S0 | 1
S
W
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E
E
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L ALC SHC RRA RRB N WRA |
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0 øøø øøø øøø øøø 0 øøø |
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1 101 1øø øøø 000 1 001 |
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1 101 1øø øøø 000 1 010 |
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0 101 1øø 001 010 1 011 |
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0 101 1øø 011 001 1 011 |
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0 101 1øø 011 010 1 011 |
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0 101 010 011 000 1 011 |
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0 101 1øø 011 000 0 øøø |
NOP
A ← inport
B ← inport
C ← A + B
C ← C + A
C ← C + B
C ← C >> 1
outport ← C
A ← inport
B ← inport
A ← A + B
outport ← A
A = R1
B = R2
Question 2:
A ← inport
B ← inport
P ← 0
mask ← 1
while A ≠ 0
temp ← A
if temp ≠
P ←
else
P ←
B ← B <<
A ← A >>
end while
outport ← P
A = R1
B = R2
P = R3
mask = R4
AND mask
0
P + B
temp = R5
P + 0
1
1
S0
start=0
S9
A=0
start≠0
S1
outport ← P
A ← inport
S8
A ← A >> 1
S7
B ← B << 1
S6
P ← P + B
A≠ 0
S2
B ← inport
temp=0
S3
P ← 0
temp≠ 0
S4
mask ← 1
S5
State Graph
temp ← A AND mask
S0
S1
S2
S3
S4
S5
S6
S7
S8
S9
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start, z≠0 | E
00 01 10 11 | N
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S0 S0 S1 S1 | 0
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S2 S2 S2 S2 | 0
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S3 S3 S3 S3 | 0
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S4 S4 S4 S4 | 0
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S5 S5 S5 S5 | 0
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S7 S6 S7 S6 | 0
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S7 S7 S7 S7 | 0
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S8 S8 S8 S8 | 0
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S9 S5 S9 S5 | 0
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S0 S0 S0 S0 | 1
S
W
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E
E
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L ALC SHC RRA RRB N WRA |
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0 øøø øøø øøø øøø 0 øøø |
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1 101 1øø øøø 000 1 001 |
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1 101 1øø øøø 000 1 010 |
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0 101 1øø 000 000 1 011 |
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0 111 1øø 000 øøø 1 100 |
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0 001 1øø 001 100 1 101 |
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0 101 1øø 011 010 1 011 |
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0 101 000 010 000 1 010 |
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0 101 010 001 000 1 001 |
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0 101 1øø 011 000 0 øøø |
NOP
A ← inport
B ← inport
P ← 0
mask ← 1
temp ← A AND mask
P ← P + B
B ← B << 1
A ← A >> 1
outport ← P