LR分析-demo2
0.LR分析
用一个栈来保存文法符号和状态的信息，一个字符串保存输入信息。
使用栈顶的状态符号和当前的输入符号来检索分析表，来决定移进-归约分析的动作。
1.样例文法
"E>E+T",
"E>T",
"T>T*F",
"T>F",
"F>(E)",
"F>id",
2.分析表(未全部列出)
3.code
//LR分析-demo2
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<stack>
#include<map>
using namespace std;
stack<string>stk, tmp, com; //stk是存入s0x1，即状态和符号的栈，tmp是用来输出的栈，
//com是在归约时确定归约式子的栈(防止错误弹出元素)
string input; //输入串
string slr[15][15]; //slr表
string action[] = { //slr表横轴动作+转换
"id","+","*","(",")","$","E","T","F"
};
//i就是id
string handle[] = { //文法
" ",
"E>E+T",
"E>T",
"T>T*F",
"T>F",
"F>(E)",
"F>id",
};
map<string, int> act, status; //转移表
map<string, int>::iterator it_act, it_trans;
string transfer[] = {
"E","T","F"
};
bool flag = false; //accept状态
int now = 0; //当前扫描字符位置
int handle_index = 0; //选中的规约式序号
void init() { //初始化，
int i = 0;
for (i = 0; i < 9;i++) {
act.insert(make_pair(action[i], i)); //建立分析表
}
int j = 0;
while (j<12) {
int i = j;
string t;
if (i < 10)
t = '0' + i;
else {
t = ('0' + (i / 10));
t += ('0' + i % 10);
}
status.insert(make_pair(t, j)); //将状态和数字对应
j++;
}
slr[0][0] = slr[4][0] = slr[6][0] = slr[7][0] = "s5"; //保存slr表
slr[1][1] = slr[8][1] = "s6";
slr[2][1] = slr[2][4] = slr[2][5] = "r2";
slr[2][2] = slr[9][2] ="s7";
slr[0][3] = slr[6][3] = slr[4][3] = slr[7][3] = "s4";
slr[1][5] = "acc";
slr[3][1] = slr[3][2] = slr[3][4] = slr[3][5] = "r4";
slr[5][1] = slr[5][2] = slr[5][4] = slr[5][5] = "r6";
slr[9][1] = slr[9][4] = slr[9][5] = "r1";
slr[8][4] = "s11";
slr[10][1] = slr[10][2] = slr[10][4] = slr[10][5] = "r3";
slr[11][1] = slr[11][2] = slr[11][4] = slr[11][5] = "r5";
slr[0][6] = "1";
slr[0][7] = slr[4][7] = "2";
slr[0][8] = slr[4][8] = slr[6][8] = "3";
slr[4][6] = "8";
slr[6][7] = "9";
slr[7][8] = "10";
}
void show() {
int count_two_char = 0; //数栈中超过两个字符的元素
while (!stk.empty()) { //用两个栈来回倒，输出字符
tmp.push(stk.top());
stk.pop();
}
while (!tmp.empty()) {
cout << tmp.top();
if (tmp.top().size()>1)
count_two_char++;
stk.push(tmp.top());
tmp.pop();
}
cout.setf(ios::right);
cout.width(11 - stk.size()- count_two_char);
cout << "|";
}
//参数1是状态，参数2是符号，符号包括终结符和非终结符，作用是找到slr表中项目
string slrFind(string stat,string ActionAndTransfer) {
string s = stat, a = ActionAndTransfer;
//cout << s + " " + a << endl;
int t1 = status[s]; //取出状态对应下标
int t2 = act[a]; //取出符号对应下标
string tmp;
if (slr[t1][t2] != "") //如果slr表中存在此项
tmp = slr[t1][t2];
else
tmp = "";
//cout << tmp << endl;
return tmp; //返回slr表中的项目
}
//参数1和2同slrFind函数，index是选中的规约式子序号
bool judge(string stat, string Transfer,int &index) {
string judg = slrFind(stat, Transfer); //得到slr表中项目
if (judg[0] != 'r') //如果这个项不是r开头，就不是归约
return false; //非归约直接返回
int i = judg[1] - '0'; //如果是归约，得到归约式子序号
index = i;
return true; //可以发起归约
}
void analysis(string s) {
string w = s;
//cout.setf(ios::right); //设置字符对其方式
//cout.width(10); //设置字符宽度
printf("----------|----------|----------\n");
printf(" 栈 | 输入 | 动作 \n");
printf("----------|----------|----------\n");
while (!flag) { //处于非接受状态
now = 0; //正在处理的输入串中的字符
if (stk.empty()) { //一开始栈为空，直接移进符号
stk.push("0");
cout << "0 |";
cout.setf(ios::right); //设置字符对其方式
cout.width(10); //设置字符宽度
cout << w;
cout << "|移进" << endl;
printf("----------|----------|----------\n");
string t1, t2;
t1 = stk.top();
if (w[now] == 'i') { //移进符号为id
t2 = "id";
now = 2;
}
else { //移进符号不为id
t2 = w[now];
now = 1;
}
stk.push(t2); //将符号压入栈
w = w.substr(now, w.size() - now); //丢弃已扫描的字符
string lr = slrFind(t1, t2); //找到对应的动作
if (lr[0] == 's') //此时是移进
lr = lr.substr(1, lr.size() - 1);
stk.push(lr);
continue;
}
show();
string serach;
if(w[0]!='i') //获取输入串的开头符号
serach =w.substr(0,1);
else
serach = w.substr(0, 2);
//cout << w[0]+""<< endl; //转换字符串不能这么做
if(judge(stk.top(), serach, handle_index)) { //归约，优先级最高
cout.setf(ios::right); //设置字符对其方式
cout.width(10); //设置字符宽度
cout << w;
cout << "|";
cout.setf(ios::left);
cout.width(10);
cout << "按" + handle[handle_index] + "归约" << endl;
printf("----------|----------|----------\n");
string ttt = handle[handle_index].substr(2, handle[handle_index].size() - 2); //得到产生式右部符号
while (ttt != "") {
if (ttt[0] == 'i'){ //将产生式右部所有非终结符暂时压入一个栈中
com.push("id");
ttt=ttt.substr(2, ttt.size() - 2);
}
else
{
string t5;
t5 = ttt[0];
com.push(t5);
ttt=ttt.substr(1, ttt.size() - 1);
}
}
while (!com.empty()) { //用这个有产生式右部所有非终结符的栈和当前栈比对，确定归约式正确
stk.pop();
string cmp1 = stk.top();
if (com.top()==cmp1) {
stk.pop();
com.pop();
}
}
string t3 = handle[handle_index];
t3 = t3[0]; //得到归约式的左部符号
string t4 = slrFind(stk.top(),t3); //用此时左部符号和当前栈顶来确认下一个动作
stk.push(t3);
stk.push(t4);
continue;
}
else { //移进操作--或者acc
cout.setf(ios::right); //设置字符对其方式
cout.width(10); //设置字符宽度
cout << w;
cout << "|";
cout.setf(ios::left);
cout.width(10);
string t1, t2;
t1 = stk.top();
if (w[now] == 'i') { //移进符号为id
t2 = "id";
now = 2;
}
else { //移进符号不为id
t2 = w[now];
now = 1;
}
stk.push(t2); //将符号压入栈
w = w.substr(now, w.size() - now); //丢弃已扫描的字符
string lr = slrFind(t1, t2); //找到对应的动作
if (lr[0] == 's'){ //如果是移进操作
lr = lr.substr(1, lr.size() - 1);
cout << "移进" << endl;
}
else if (lr == "acc") { //或者是接受状态
cout << "接受" << endl;
flag = true;
}
stk.push(lr);
printf("----------|----------|----------\n");
continue;
}
}
}
int main(void) {
init();
input = "id*id+id"; //输入串
input += "$";
analysis(input);
return 0;
}
4.样例输出
5. To be continued.
(。・∀・)ノ
