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 |
| 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 |
|
|
|
|
实验程序：
1 #include<bits/stdc++.h> 2 #define ROW 12 3 #define COLUMN 9 4 using namespace std; 5 //产生式 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; //状态栈 32 stack<char> schar; //符号栈 33 struct Node{ 34 char type; 35 int num; 36 }; 37 //打印步骤 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 }