武汉大学计算机学院《编译原理》06-07学年试题参考答案

S1. true := newlabel(); S1. false := S. false; S2. true := S. true; S2. false := S. false; S. code := S1. code gen(S1. true, ":") S2. code; S1. element := S. element; S2. element := S. element; S1. true := S. true; S1. false := newlabel(); S2. true := S. true; S2. false := S. false; S. code := S1. code gen(S1. false, ":") S2. code; S1. element := S. element; S2. element := S. element; S1. true := newlabel(); S1. false := S. false; S2. true := S. false; S2. false := S. true; S. code := S1. code gen(S1. true, ":") S2. code; S1. element := S. element; S2. element := S. element; S1. true := S. false; S1. false := S. true; S. code := S1. code; S1. element := S. element; S1. true := S. true; S1. false := S. false; S. code := S1. code; S1. element := S. element;
(3) LL(1)
Select(S → AB) = { a }; Select(S → CDa) =
{ c, d, a }, Select(S → AB) ∩ Select(S → CDa) = ∅
(1)
“a ∩ a ∪ a”
S =⇒ S ∩ S
lm
=⇒ a ∩ S
lm
=⇒ a ∩ S ∪ S
lm
=⇒ a ∩ a ∪ S
a$
shift
(11) I0SI13 − I7aI12
$ reduce S → a
(12) I0SI13 − I7SI8
$ reduce S → S − S
(13) I0SI13
$
accept
3
(1)
E → id ∈ S S. true := E. true; S. false := E. false;
(3) call by reference 24 42
(4) call by value-result 21 42
4
main()
frame pointer
1 address memory note
...... x fp x-4 ret add x-8 0 x-9 ’7’ x-10 ’6’ x-11 ’5’ x-12 ’4’ x-13 ’3’ x-14 ’2’ x-15 ’1’ ← t x-16 0 x-17 ’c’ x-18 ’b’ x-19 ’a’ ← s
(2)
(1) if x in A goto L2
(2) goto L1
(3) L2: if x in B goto L1
(4) goto Lfalse
(3) L1: if x in C goto Lfalse
(4) goto Ltrue
(1) call by name 43 34
(2) call by value 21 12
2004
2006-2007
(1) 0(0|1) ∗ 11 (2)
start
ε ε
ε
ε
0
ε
ε
ε
1
ε
ε
(3) start
ε
0
1
0
0
1
0
1
00 ε
1
1
ε
(4)
b
b
a
start
a
0,2
1,4
3ቤተ መጻሕፍቲ ባይዱ5
b
(1) S → aSbb abb; (2) a b
a
b;
(ab|b)+
1
(3) T =⇒ T ab
lm
=⇒ T bab
......
2 address memory note
...... x ’6’ x-4 ’5’ x-8 ’4’ x-9 ’3’ x-10 ’2’ x-11 ’1’ ← s x-12 0 x-13 ’7’ x-14 ’6’ x-15 ’5’ x-16 4 x-17 ’3’ x-18 ’2’ x-19 ’1’ ← t
S. element := id. name; E. code := S. code;
S→a
S. code := gen("if", S. element, "in", a. name, "goto", S. true) gen("goto", S. false);
S → S1 ∩ S2 S → S1 ∪ S2 S → S1 − S2 S → −S1 S → (S1)
(5) I0SI13
∪a − a$
shift
(6) I0SI13 ∪ I5
a − a$
shift
(7) I0SI13 ∪ I5aI12
−a$ reduce S → a
(8) I0SI13 ∪ I5SI6
−a$ reduce S → S ∪ S
(9) I0SI13
−a$
shift
(10) I0SI13 − I7
13 s3 s5 s7 / / / acc /
(3)
stack
input
action
(1) I0
−a ∪ a − a$
shift
(2) I0 − I1
a ∪ a − a$
shift
(3) I0 − I1aI12
∪a − a$ reduce S → a
(4) I0 − I1SI2
∪a − a$ reduce S → −S
......
x, main()
AR
2 address memory note
...... x fp x-4 ret add x-8 0 x-9 ’c’ x-10 ’b’ x-11 ’a’ ← s x-12 0 x-13 ’7’ x-14 ’6’ x-15 ’5’ x-16 4 x-17 ’3’ x-18 ’2’ x-19 ’1’ ← t
(2)
state
action
goto
∩∪− ( ) a $ S
0 / / s1 s9 / s12 / 13
1 / / s1 s9 / s12 / 2
2 r4 r4 r4 / r4 / r4 /
3 / / s1 s9 / s12 / 4
4 r1 r1 r1 / r1 / r1 /
5 / / s1 s9 / s12 / 6
6 s3 r2 r2 / r2 / r2 /
7 / / s1 s9 / s12 / 8
8 s3 s5 s7 / r3 / r3 /
9 / / s1 s9 / s12 / 10
10 s3 s5 s7 / s11 / / /
11 r5 r5 r5 / r5 / r5 /
12 r6 r6 r6 / r6 / r6 /
lm
=⇒ a ∩ a ∪ a
lm
S =⇒ S ∪ S
lm
=⇒ S ∩ S ∪ S
lm
=⇒ a ∩ S ∪ S
lm
=⇒ a ∩ a ∪ S
lm
=⇒ a ∩ a ∪ a
lm
2
(2) S → D−S D D → D∪U U U → U ∩X X X → −X A A → (S) a
(1) I5 = S → S ∪ •S, S → •S ∩ S, S → •S ∪ S, S → •S − S, S → • − S, S → •(S), S → •a I10 = S → (S•), S → S • ∩S, S → S • ∪S, S → S • −S
......
“strcpy(s, t);”
main() AR
1 address memory note
...... x fp x-4 ret add x-8 0 x-9 ’7’ x-10 ’6’ x-11 ’5’ x-12 0 x-13 ’7’ x-14 ’6’ x-15 ’5’ ← t x-16 ’4’ x-17 ’3’ x-18 ’2’ x-19 ’1’ ← s
......
1 “strcpy(s, t);”
“567”;
2
t
“567\0567\0”,
s
“1234567\0”,
5
lm
=⇒ T bbab
lm
=⇒ abbbab
lm
T
Tab
T
b
T
b
ab
(1) First(S) = { a, c, d }; First(A) = { a }; First(B) = { b, ε }; First(C) = { c, }; First(D) = { d, ε };
(2) Follow(S) = { $ }; Follow(A) = { b }; Follow(B) = { $ }; Follow(C) = { a, d }; Follow(D) = { a };