SLR(1)分析法
由于LR(0)的能力實(shí)在是太弱了。例如：
I = { X=>α·bβ，
A=>α·，
B=>α· }
這時(shí)候就存在兩個(gè)沖突。
1、移進(jìn)和規(guī)約的沖突；
2、規(guī)約和規(guī)約的沖突。
SLR（1）就是為了解決沖突而設(shè)計(jì)的，解決沖突的方法就是向后多看一個(gè)字符，這就是SLR（1）。
簡(jiǎn)而言之就是為每個(gè)非終結(jié)符，計(jì)算出它們的follow集。從而可以解決移進(jìn)與規(guī)約、規(guī)約與規(guī)約的沖突了。
SLR（1）所說的多看一個(gè)字符在構(gòu)造分析表的時(shí)候就根據(jù)follow集已經(jīng)構(gòu)造好了，分析程序和LR(0)是一樣的，分析表不同。
具體實(shí)現(xiàn)如下：
拓廣文法 G'[S']:
(0) S'→S
(1) S→ABC
(2) A→Aa
(3) A→a
(4) B→Bb
(5) B→b
(6) C→Cc
(7) C→c
1、計(jì)算初始狀態(tài)的項(xiàng)目集
Q0=CLOSURE({S'→?S })={ S'→?S , S→?ABC, A→?Aa, A→?a };
2、計(jì)算每個(gè)狀態(tài)的項(xiàng)目集
Q1=GO(Q0,a)=CLOSURE({A→a ?})={ A→a? };
Q2=GO(Q0,S)=CLOSURE({S'→S ?})={ S'→S ? };
Q3=GO(Q0,A) = CLOSURE({S→A?BC, A→A?a}) = {S→A?BC, A→A?a, B→?Bb, B→?b}; Q4=GO(Q3,a)=CLOSURE({A→Aa? })={ A→Aa? };
Q5=GO(Q3,b)=CLOSURE({B→b? })={ B→b?};
Q6=GO(Q3,B)=CLOSURE({S→AB?C, B→B?b }) ={ S→AB?C, B→B?b , C→?Cc , C→?c }; Q7=GO(Q6,b)=CLOSURE({B→Bb ?})={ B→Bb ?};
Q8=GO(Q6,c)=CLOSURE({C→c ?})={ C→c ?};
Q9=GO(Q6,C)=CLOSURE({S→ABC?, C→C?c })={ S→ABC?, C→C?c }; Q10=GO(Q9,c)=CLOSURE({C→Cc? })={ C→Cc?};
3、構(gòu)造識(shí)別可歸約前綴的 DFA
4、計(jì)算文法的 FIRST 和 FOLLOW 集合
|
非終結(jié)符 |
FIRST |
FOLLOW |
|
S |
a |
# |
|
A |
a |
a,b |
|
B |
b |
b,c |
|
C |
c |
c,# |
狀態(tài)節(jié)點(diǎn) Q9= { S→ABC?, C→C?c }中存在存在移進(jìn)-規(guī)約沖突。
{b}∩FOLLOW(S) ={b}∩{#}=Φ，因此文法是 SLR(1)文法。
5、構(gòu)造 SLR(1)分析表
|
a |
b |
c |
# |
S |
A |
B |
C |
|
|
0 |
S1 |
2 |
3 |
|||||
|
1 |
R3 |
R3 |
||||||
|
2 |
acc |
|||||||
|
3 |
S4 |
S5 |
6 |
|||||
|
4 |
R2 |
R2 |
||||||
|
5 |
R5 |
R5 |
||||||
|
6 |
S7 |
S8 |
9 |
|||||
|
7 |
R4 |
R4 |
||||||
|
8 |
R7 |
R7 |
||||||
|
9 |
S10 |
R1 |
||||||
|
10 |
R6 |
R6 |
實(shí)驗(yàn)程序：
1 #include<bits/stdc++.h>
2 #define ROW 12
3 #define COLUMN 9
4 using namespace std;
5 //產(chǎn)生式
6 string products[8][2]={
7 {"S'","S"},
8 {"S","ABC"},
9 {"A","Aa"},
10 {"A","a"},
11 {"B","Bb"},
12 {"B","b"},
13 {"C","Cc"},
14 {"C","c"}
15 };
16 //SLR（1）分析表
17 string actiontable[ROW][COLUMN]={
18 {"","a","b","c","#","S","A","B","C"},
19 {"0","s1","","","","2","3","",""},
20 {"1","r3","r3","","","","","",""},
21 {"2","","","","acc","","","",""},
22 {"3","s4","s5","","","","","6",""},
23 {"4","r2","r2","","","","","",""},
24 {"5","","r5","r5","","","","",""},
25 {"6","","s7","s8","","","","","9"},
26 {"7","","r4","r4","","","","",""},
27 {"8","","","r7","r7","","","",""},
28 {"9","","","s10","r1","","","",""},
29 {"10","","","r6","r6","","","",""}
30 };
31 stack<int> sstatus; //狀態(tài)棧
32 stack<char> schar; //符號(hào)棧
33 struct Node{
34 char type;
35 int num;
36 };
37 //打印步驟
38 void print_step(int times){
39 stack<char> tmp2;
40 cout<<times<<setw(4);
41 while(!schar.empty()){
42 char t=schar.top();
43 schar.pop();
44 tmp2.push(t);
45 cout<<t;
46 }
47 while(!tmp2.empty()){
48 int t=tmp2.top();
49 tmp2.pop();
50 schar.push(t);
51 }
52 }
53 //查表
54 Node Action_Goto_Table(int status,char a){
55 int row=status+1;
56 string tmp;
57 for(int j=1;j<COLUMN;j++){
58 if(a==actiontable[0][j][0]){
59 tmp=actiontable[row][j];
60 }
61 }
62 Node ans;
63 if(tmp[0]>='0'&&tmp[0]<='9'){
64 int val=0;
65 for(int i=0;i<tmp.length();i++){
66 val=val*10+(tmp[i]-'0');
67 }
68 ans.num=val;
69 ans.type=' ';
70 }else if(tmp[0]=='s'){
71 int val=0;
72 for(int i=1;i<tmp.length();i++){
73 val=val*10+(tmp[i]-'0');
74 }
75 ans.type='s';
76 ans.num=val;
77 }else if(tmp[0]=='r'){
78 int val=0;
79 for(int i=1;i<tmp.length();i++){
80 val=val*10+(tmp[i]-'0');
81 }
82 ans.type='r';
83 ans.num=val;
84 }else if(tmp[0]=='a'){
85 ans.type='a';
86 }else{
87 ans.type=' ';
88 }
89 return ans;
90 }
91 //SLR(1)分析算法
92 bool SLR1(string input){
93 while(!sstatus.empty()){
94 sstatus.pop();
95 }
96 while(!schar.empty()){
97 schar.pop();
98 }
99 int times=0;
100 bool flag=true;
101 int st=0;
102 sstatus.push(st);
103 schar.push('#');
104 int i=0;
105 char a=input[i];
106 while(true){
107 Node action=Action_Goto_Table(st,a);
108 if(action.type=='s'){
109 st=action.num;
110 sstatus.push(st);
111 schar.push(a);
112 a=input[++i];
113 print_step(++times);
114 cout<<setw(10)<<'s'<<st<<endl;
115
116 }else if(action.type=='r'){
117 int n=action.num;
118 string ls=products[n][0];
119 string rs=products[n][1];
120 for(int j=0;j<rs.length();j++){
121 sstatus.pop();
122 schar.pop();
123 }
124 schar.push(ls[0]);
125 st=sstatus.top();
126 action =Action_Goto_Table(st,ls[0]);
127 st=action.num;
128 sstatus.push(st);
129 print_step(++times);
130 cout<<setw(10)<<'r'<<" "<<ls<<"->"<<rs<<endl;
131
132 }else if(action.type=='a'){
133 flag=true;
134 break;
135 }else{
136 flag=false;
137 break;
138 }
139 }
140 return flag;
141 }
142 int main(){
143 string input;
144 while(cin>>input){
145 if(SLR1(input)){
146 cout<<"syntax correct"<<endl;
147 }else{
148 cout<<"syntax error"<<endl;
149 }
150 }
151 return 0;
152 }
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