•  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。

• 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

• 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;
}

• 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：