实验任务：

实现LR语法分析器

实验要求：

根据编译原理理论课教材中例3.22给出的算术表达式文法以及该文法的LR分析表，用C语言编写接受算术表达式为输入的语法分析器，以控制台（或文本文件，也可以结合词法分析器完成）为输入，控制台（或文件）输出产生式序列形式的分析结果。

实验内容：

教材中例3.22给出的算术表达式文法只有+、运算，对应的LR分析表也比较单一，为了编写接受算术表达式为输入的语法分析器，我将文法完善为下列文法，然后先在纸上写出LR(0)项目集规范族来构造LR分析表，再将其写入程序作为该文法的LR分析表。然后再依据使用分析表对输入符号串做分析的过程编写程序，使得用LR分析法对任意输入的符号串进行分析。
（1）E-> E+T
（2）E- >E-T
（3）E- >T
（3）T- >TF
（4）T- >T/F
（5）T- >F
（6）F- >(E)
（7）F- >i

主要代码:

运行示例：

实验小结：

1.在看教材中的例3.22时，对于一个算术表达式而言绝不止+、*符号，故例3.22的文法具有一定的局限性，所以我们可以将文法进行拓展，即终结符有±*/，涵盖算术表示式应有的所有运算符，在此基础上我们还需要构造LR(0)项目集规范族，以构造出LR分析表，在确定正确的情况下将其输入程序中以作分析。

2.与上一个实验构造LL(1)语法分析器作对比，有很多的不同之处，LL(1)文法与LR(0)或LR(1)文法的分析技术就是不同的，所以本实验在汲取上个实验部分相同点的基础上需要改变分析思路，这样才能构造出正确的分析过程。

源程序

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<stack>
using namespace std;///					   0		1		2		3		4		5		6		7		8		9		10
///表格数组                +       -        *       /       i      （      ）       # 	    E	    T		 F
char LR0[50][50][100] = {{"null" ,"null" ,"null", "null" ,"S4"   ,"S5"   ,"null" ,"null" ,"1"    ,"2"    ,"3"   },///   0{"S6"   ,"S7"   ,"null", "null" ,"null ","null" ,"null" ,"acc"  ,"null" ,"null" ,"null"},///   1{"r3"   ,"r3"   ,"S8"  , "S9"   ,"null" ,"null" ,"r3"   ,"r3"   ,"null" ,"null" ,"null"},///   2{"r6"   ,"r6"   ,"r6"  , "r6"   ,"null" ,"null" ,"r6"   ,"r6"   ,"null" ,"null" ,"null"},///   3{"r8"   ,"r8"   ,"r8"  , "r8"   ,"null" ,"null" ,"r8"   ,"r8"   ,"null" ,"null" ,"null"},///   4{"null" ,"null" ,"null", "null" ,"S4"   ,"S5"   ,"null" ,"null" ,"10"   ,"2"    ,"3"   },///   5{"null" ,"null" ,"null", "null" ,"S4"   ,"S5"   ,"null" ,"null" ,"null" ,"11"   ,"3"   },///   6{"null" ,"null" ,"null", "null" ,"S4"   ,"S5"   ,"null" ,"null" ,"null" ,"12"   ,"3"   },///   7{"null" ,"null" ,"null", "null" ,"S4"   ,"S5"   ,"null" ,"null" ,"null" ,"null" ,"13"  },///   8{"null" ,"null" ,"null", "null" ,"S4"   ,"S5"   ,"null" ,"null" ,"null" ,"null" ,"14"  },///   9{"S6"   ,"S7"   ,"null", "null" ,"null" ,"null" ,"S15"  ,"null" ,"null" ,"null" ,"null"},///   10{"r1"   ,"r1"   ,"S8"  , "S9"   ,"null" ,"null" ,"r1"   ,"r1"   ,"null" ,"null" ,"null"},///   11{"r2"   ,"r2"   ,"S8"  , "S9"   ,"null" ,"null" ,"r2"   ,"r2"   ,"null" ,"null" ,"null"},///   12{"r4"   ,"r4"   ,"r4"  , "r4"   ,"null" ,"null" ,"r4"   ,"r4"   ,"null" ,"null" ,"null"},///   13{"r5"   ,"r5"   ,"r5"  , "r5"   ,"null" ,"null" ,"r5"   ,"r5"   ,"null" ,"null" ,"null"},///   14{"r7"   ,"r7"   ,"r7"  , "r7"   ,"null" ,"null" ,"r7"   ,"r7"   ,"null" ,"null" ,"null"},///   15};
char L[200]="+-*/i()#ETF";    ///列判断依据
int  del[10]={0,3,3,1,3,3,1,3,1};//0-6号文法每个文法长度
char head[20]={'S','E','E','E','T','T','T','F','F'};
stack<int>con;    ///状态栈
stack<char>cmp;   ///符号栈
char cod[300]="0";///初始状态栈对应输出数组
int cindex = 0;
char sti[300]="#";///初始符号栈对应输出数组
int sindex = 0;
int findL(char b)///对应列寻找
{for(int i = 0; i <= 10; i++){if(b==L[i]){return i;}}return -1;
}
void error(int x, int y)       ///报错输出
{printf("第%d行%c列为空!",x,L[y]);
}int calculate(int l, char s[])
{int num = 0;for(int i = 1; i < l; i ++){num =  num*10+(s[i]-'0');}return num;
}
void analyze(char str[],int len)///分析主体过程
{int cnt = 1;printf("步骤		状态栈		符号栈    输入串    ACTION    GOTO\n");int LR = 0;while(LR<=len){printf("(%d)		%-10s	%-10s",cnt,cod,sti);///步骤，状态栈，符号栈输出cnt++;for(int i = LR; i < len; i++)///输入串输出{printf("%c",str[i]);}for(int i = len-LR; i<10;i++)printf(" ");int x = con.top();///状态栈栈顶int y = findL(str[LR]);///待判断串串首if(strcmp(LR0[x][y],"null")!=0){int l = strlen(LR0[x][y]);///当前Ri或Si的长度if(LR0[x][y][0]=='a')///acc{printf("acc        \n");///ACTION与GOTOreturn ;}else if(LR0[x][y][0]=='S')///Si{printf("%-10s \n",LR0[x][y]);///ACTION与GOTOint t = calculate(l,LR0[x][y]);///整数con.push(t);sindex++;sti[sindex] = str[LR];cmp.push(str[LR]);if(t<10){cindex++;cod[cindex] = LR0[x][y][1];}else{int k = 1;cindex++;cod[cindex] = '(';while(k<l){cindex++;cod[cindex] = LR0[x][y][k];k++;}cindex++;cod[cindex] = ')';}LR++;}else if(LR0[x][y][0]=='r')//ri,退栈，ACTION和GOTO{printf("%-10s",LR0[x][y]);int t = calculate(l,LR0[x][y]);int g = del[t];while(g--){con.pop();cmp.pop();sti[sindex] = '\0';sindex--;}g = del[t];while(g>0){if(cod[cindex]==')'){cod[cindex]='\0';cindex--;for(;;){if(cod[cindex]=='('){cod[cindex]='\0';cindex--;break;}else{cod[cindex]='\0';cindex--;}}g--;}else{cod[cindex] = '\0';cindex--;g--;}}///cmp.push(head[t]);sindex++;sti[sindex] = head[t];x = con.top();y = findL(cmp.top());if(x==5&&y==8){t = 10;cindex++;cod[cindex] = '(';cindex++;cod[cindex] = '1';cindex++;cod[cindex] = '0';cindex++;cod[cindex] = ')';}else if(x==6&&y==9){t = 11;cindex++;cod[cindex] = '(';cindex++;cod[cindex] = '1';cindex++;cod[cindex] = '1';cindex++;cod[cindex] = ')';}else if(x==7&&y==9){t=12;cindex++;cod[cindex] = '(';cindex++;cod[cindex] = '1';cindex++;cod[cindex] = '2';cindex++;cod[cindex] = ')';}else if(x==8&&y==10){t=13;cindex++;cod[cindex] = '(';cindex++;cod[cindex] = '1';cindex++;cod[cindex] = '3';cindex++;cod[cindex] = ')';}else if(x==9&&y==10){t=14;cindex++;cod[cindex] = '(';cindex++;cod[cindex] = '1';cindex++;cod[cindex] = '1';cindex++;cod[cindex] = '4';}else{t = LR0[x][y][0]-'0';cindex++;cod[cindex] = LR0[x][y][0];}con.push(t);printf("%-10d\n",t);}else{int t = LR0[x][y][0]-'0';char ch = ' ';printf("%-10c%-10d\n",ch,t);con.push(t);cindex++;cod[cindex] = LR0[x][y][0];sindex++;sti[sindex] = 'E';LR++;}}else{error(x,y);return ;///报错}}}
void chart()///测试表函数
{printf("-\t+\t-\t*\t/\ti\t(\t)\t#\tE\tT\tF\n");for(int i = 0; i <= 15; i++){printf("%d",i);for(int j = 0 ; j <= 11; j++)printf("\t%s",LR0[i][j]);cout<<endl;}cout<<endl;
}
int main()
{chart();con.push(0);cmp.push('#');char str[200];///输入串cout<<"请输入字符串（带#）："<<endl;cin>>str;int len = strlen(str);//输入串长度 analyze(str,len);return 0;
}