SLR（1）分析法

SLR（1）分析法
由于LR(0)的能⼒实在是太弱了。

例如：
I = { X=>α·bβ，
A=>α·，
B=>α· }
这时候就存在两个冲突。

1、移进和规约的冲突；
2、规约和规约的冲突。

SLR（1）就是为了解决冲突⽽设计的，解决冲突的⽅法就是向后多看⼀个字符，这就是SLR（1）。

简⽽⾔之就是为每个⾮终结符，计算出它们的follow集。

从⽽可以解决移进与规约、规约与规约的冲突了。

SLR（1）所说的多看⼀个字符在构造分析表的时候就根据follow集已经构造好了，分析程序和LR(0)是⼀样的，分析表不同。

具体实现如下：
拓⼴⽂法 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、计算初始状态的项⽬集
Q0=CLOSURE({S'→•S })={ S'→•S , S→•ABC, A→•Aa, A→•a };
2、计算每个状态的项⽬集
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、构造识别可归约前缀的 DFA
4、计算⽂法的 FIRST 和 FOLLOW 集合
⾮终结符FIRST FOLLOW
S a#
A a a,b
B b b,c
C c c,#
状态节点 Q9= { S→ABC•, C→C•c }中存在存在移进-规约冲突。

{b}∩FOLLOW(S) ={b}∩{#}=Φ，因此⽂法是 SLR(1)⽂法。

5、构造 SLR(1)分析表
a b c#S A B C 0S123
1R3R3
2acc
3S4S56
4R2R2
5R5R5
6S7S89 7R4R4
8R7R7
9S10R1
10R6R6
实验程序：
1 #include<bits/stdc++.h>
2#define ROW 12
3#define COLUMN 9
4using namespace std;
5//产⽣式
6string 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
17string 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; //状态栈
32 stack<char> schar; //符号栈
33struct Node{
34char type;
35int num;
36 };
37//打印步骤
38void print_step(int times){
39 stack<char> tmp2;
40 cout<<times<<setw(4);
41while(!schar.empty()){
42char t=schar.top();
43 schar.pop();
44 tmp2.push(t);
45 cout<<t;
46 }
47while(!tmp2.empty()){
48int t=tmp2.top();
49 tmp2.pop();
50 schar.push(t);
51 }
52 }
53//查表
54 Node Action_Goto_Table(int status,char a){ 55int row=status+1;
56string tmp;
57for(int j=1;j<COLUMN;j++){
58if(a==actiontable[0][j][0]){
59 tmp=actiontable[row][j];
60 }
61 }
62 Node ans;
63if(tmp[0]>='0'&&tmp[0]<='9'){
64int val=0;
65for(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'){
71int val=0;
72for(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'){
78int val=0;
79for(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 }
89return ans;
90 }
91//SLR(1)分析算法
92bool SLR1(string input){
93while(!sstatus.empty()){
94 sstatus.pop();
95 }
96while(!schar.empty()){
97 schar.pop();
98 }
99int times=0;
100bool flag=true;
101int st=0;
102 sstatus.push(st);
103 schar.push('#');
104int i=0;
105char a=input[i];
106while(true){
107 Node action=Action_Goto_Table(st,a); 108if(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'){
117int n=action.num;
118string ls=products[n][0];
119string rs=products[n][1];
120for(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;
134break;
135 }else{
136 flag=false;
137break;
138 }
139 }
140return flag;
141 }
142int main(){
143string input;
144while(cin>>input){
145if(SLR1(input)){
146 cout<<"syntax correct"<<endl;
147 }else{
148 cout<<"syntax error"<<endl;
149 }
150 }
151return0;
152 }。