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LR(1)分析法实验设计思想及算法
（1）若ACTION[sm , ai] = s则将s移进状态栈，并把输入符号加入符号栈，则三元式变成
为：(s0s1…sm s , #X1X2…Xm ai , ai+1…an#)
（2） 若ACTION[sm , ai] = rj则将第j个产生式A->β进行归约。此时三元式变为
(s0s1…sm-r s , #X1X2…Xm-rA , aiai+1…an#)
（3） 若ACTION[sm , ai]为“接收”，则三元式不再变化，变化过程终止，宣布分析成功。
（4） 若ACTION[sm , ai]为“报错”，则三元式的变化过程终止，报告错误。
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测试所用文法及其分析表
（1） E-> E+T
（2） E->T
（3） T-> T*F
（4） T->F
（5） F-> (E)
（6） F-> i
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代码实现
#ifndef _LR_
#define _LR_
using namespace std;
class Grammar{
public:
//产生式的个数
int grammarNum ;
//定义产生式数组
string formula[100] = {" ","E->E+T","E->T","T->T*F","T->F","F->(E)","F->i"};
Grammar(){
grammarNum = 6;
}
};
//定义LR文法的分析表
class LRAnalyseTable{
public:
char terminalChar[100]={'i','+','*','(',')','#'};
//定义终结符的个数
int terNum =6;
char nonTerminalChar[100]={'E','T','F'};
//定义非终结符的个数
int nonTerNum = 3;
//定义状态数
int statusNum = 12;
string action[12][6]={{"s5","","","s4","",""},{"","s6","","","","acc"},{"","r2","s7","","r2","r2"},{"","r4","r4","","r4","r4"},{"s5","","","s4","",""},{"","r6","r6","","r6","r6"},{"s5","","","s4","",""}
,{"s5","","","s4","",""},{"","s6","","","s11",""},{"","r1","s7","","r1","r1"},{"","r3","r3","","r3","r3"},{"","r5","r5","","r5","r5"}};
int goTo[12][3] = {{1,2,3},{-1,-1,-1},{-1,-1,-1},{-1,-1,-1},{8,2,3},{-1,-1,-1},{-1,9,3},{-1,-1,10},{-1,-1,-1},{-1,-1,-1},{-1,-1,-1},{-1,-1,-1}};
//获取终结符的索引
int getTerminalIndex(char var){
for(int i=0;i<terNum;i++){
if(terminalChar[i] == var){
return i;
}
}
return -1;
}
//获取非终结符的索引
int getNonTerminalIndex(char var){
for(int i=0;i<nonTerNum;i++){
if(nonTerminalChar[i] == var){
return i;
}
}
return -1;
}
};
#endif // _LR_
#include <iostream>
#include <vector>
#include <iomanip>
#include <cstring>
#include <sstream>
#include "LR.h"
using namespace std;
//定义状态栈
vector<int> status;
//定义符号栈
vector<char> sign;
//定义输入的字符串
vector<char> inputStr;
//定义文法
Grammar grammar;
//定义LR分析表
LRAnalyseTable analyseTable;
//读取输入的字符串
void readStr();
//对栈容器进行输出,i=0,返回status中的字符串,i=1,返回sign中的字符串，i=2返回inputStr
string vectTrancStr(int i);
//总控，对输入的字符串进行分析
void LRAnalyse();
int main()
{
readStr();
LRAnalyse();
return 0;
}
//读取输入的字符串
void readStr(){
char ch;
cout<<"请输入分析的字符串：";
cin>>ch;
while( ch != '#'){
inputStr.push_back(ch);
cin>>ch;
}
//把#加入容器
inputStr.push_back('#');
}
//对栈容器进行输出,i=0,返回status中的字符串,i=1,返回sign中的字符串，i=2返回inputStr中的字符串
string vectTrancStr(int i){
char buf[100];
int count = 0;
//输出状态栈
if(i == 0){
vector<int>::iterator it =status.begin();
//将数字转化为字符串
string str,tempStr;
for(it;it!= status.end();it++){
stringstream ss;
ss << *it;
ss >> tempStr;
str+=tempStr;
}
return str;
}
//输出符号栈
else if(i == 1){
vector<char>::iterator it = sign.begin();
for(it ; it != sign.end() ;it++){
buf[count] = *it;
count++;
}
}
//输出待分析的字符串
else{
vector<char>::iterator it = inputStr.begin();
for(it ; it != inputStr.end();it++){
buf[count] = *it;
count++;
}
}
buf[count] = '\0';
string str(buf);
return str;
}
//总控，对输入的字符串进行分析
void LRAnalyse(){
//步骤
int step = 1;
//把状态0入栈
status.push_back(0);
//把#加入符号栈
sign.push_back('#');
//输出初始栈状态
cout<<setw(10)<<"步骤"<<setw(10)<<"状态栈"<<setw(10)<<"符号栈"<<setw(10)<<"输入串"<<setw(25)<<"动作说明"<<endl;
//初始状态
int s =0;
//保存之前的状态
int oldStatus;
//获取初始符号
char ch = inputStr.front();
//如果action[s][ch] =="acc" ，则分析成功
while(analyseTable.action[s][analyseTable.getTerminalIndex(ch)] != "acc"){
//获取字符串
string str = analyseTable.action[s][analyseTable.getTerminalIndex(ch)];
//如果str为空，报错并返回
if(str.size() == 0){
cout<<"出错";
return ;
}
//获取r或s后面的数字
stringstream ss;
ss << str.substr(1);
ss >> s;
//如果是移进
if(str.substr(0,1) == "s"){
cout<<setw(10)<<step<<setw(10)<<vectTrancStr(0)<<setw(10)<<vectTrancStr(1)<<setw(10)<<vectTrancStr(2)<<setw(10)<<"A"<<"CTION["<<status.back()<<","<<ch<<"]=S"<<s<<","<<"状态"<<s<<"入栈"<<endl;
//输入符号入栈
sign.push_back(ch);
inputStr.erase(inputStr.begin());
//将状态数字入栈
status.push_back(s);
}
//如果是归约
else if(str.substr(0,1) == "r"){
//获取第S个产生式
string formu = grammar.formula[s];
//cout<<s<<endl;
int strSize = formu.size();
//将产生式转化为字符数组
char buf[100];
strcpy(buf,formu.c_str());
//获取产生式的首字符
char nonTerCh = buf[0];
//获取符号栈的出栈次数
int popCount = strSize - 3;
//反向迭代
vector<int>::reverse_iterator rit = status.rbegin();
int i= 0;
for(rit;rit != status.rend();rit++){
i++;
if(i == popCount+1){
oldStatus = * rit;
break;
}
}
int r = s;
//修改s
s = analyseTable.goTo[oldStatus][analyseTable.getNonTerminalIndex(nonTerCh)];
cout<<setw(10)<<step<<setw(10)<<vectTrancStr(0)<<setw(10)<<vectTrancStr(1)<<setw(10)<<vectTrancStr(2)<<setw(10)<<"r"<<r<<(string)":"+grammar.formula[r]+(string)"归约,GOTO{"<<oldStatus<<","<<nonTerCh<<")="<<s<<"入栈"<<endl;
//对符号栈进行出栈和状态栈进行出栈
for(int i=0 ;i< popCount;i++){
sign.pop_back();
status.pop_back();
}
//再对产生式的开始符号入栈
sign.push_back(nonTerCh);
//再把新的状态入栈
status.push_back(s);
}
else{
//什么都不处理
}
//步骤数加1
step++;
//获取栈顶状态
s = status.back();
//获取输入的字符
ch = inputStr.front();
}
cout<<setw(10)<<step<<setw(10)<<vectTrancStr(0)<<setw(10)<<vectTrancStr(1)<<setw(10)<<vectTrancStr(2)<<setw(10)<<"A"<<"cc:分析成功"<<endl;
}
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实验结果
对于正确的输入串i+i*i运行结果如下：
对于不正确的输入串i+i)运行结果如下：
