---恢复内容开始---
这是一份编译原理实验报告，分析表是手动造的，可以作为借鉴。
基于 SLR(1) 分析法的语法制导翻译及中间代码生成程序设计原理与实现
1 、理论传授
语法制导的基本概念，目标代码结构分析的基本方法，赋值语句语法制导生成四元式的
基本原理和方法，该过程包括语法分析和语义分析过程。
2 、目标任务
[ 实验 项目] 完成以下描述赋值语句 SLR(1)文法语法制导生成中间代码四元式的过程。
G[A]:A→V=E
E→E+T∣E-T∣T
T→T*F∣T/F∣F
F→(E)∣i
V→i
[ 设计说明 ] 终结符号 i 为用户定义的简单变量，即标识符的定义。
[ 设计要求]
（1）构造文法的 SLR(1)分析表，设计语法制导翻译过程，给出每一产生式
对应的语义动作；
（2）设计中间代码四元式的结构；
（3）输入串应是词法分析的输出二元式序列，即某赋值语句“专题 1”的输出结果，输出为赋值语句的四元式序列中间文件；
（4）设计两个测试用例（尽可能完备），并给出程序执行结果四元式序列。
3、 程序功能描述
在第一次实验词法分析输出结果的基础上设计SLR1文法分析过程，并了解四元式的形成：
- 输入串为实验一的二元式序列
- 输出为对输入串的SLR(1)文法的判断结果
- 输出有针对输入串的SLR（1）文法的具体分析过程
- 有对输入串的四元式输出序列
4、 主要数据结构描述
二元组结构体，用来存储词法分析程序输出的二元组对 <类别，单词>：
int count;
struct eryuanzu
{
int a;
char temp[COUNT];
}m[COUNT];
void out(int a,char* temp){// 打印二元组
printf("< %d %s >\n",a,temp);
m[count].a=a;
strcpy(m[count].temp,temp); //
count++;
}
SLR1分析过程中所要用到的状态栈、符号栈等：
stack<int> state; //状态栈
stack<char> sign; //符号栈
char st; //规约弹出时，状态栈顶元素
int flag=0; //标志是否是SLR
stack<string> place; //变量地址栈
ACTION表，二维数组表示：
/* i ( ) + - * / = #
以1开头的百位数为s移进项，0为error，-1为accept，其余的一位两位数是r规约项*/
int ACTION[20][9]={{103,0,0,0,0,0,0,0,0},//0
{0,0,0,0,0,0,0,0,-1},
{0,0,0,0,0,0,0,104,0},
{0,0,0,0,0,0,0,10,0},
{109,108,0,0,0,0,0,0},
{0,0,0,110,111,0,0,0,1},//5 mnl;;huhyhjhjio
{0,0,4,4,4,112,113,0,4},
{0,0,7,7,7,7,7,0,7},
{109,108,0,0,0,0,0,0,0},
{0,0,9,9,9,9,9,0,9},
{109,108,0,0,0,0,0,0,0},//10
{109,108,0,0,0,0,0,0,0},
{109,108,0,0,0,0,0,0,0},
{109,108,0,0,0,0,0,0,0},
{0,0,119,110,111,0,0,0,0},
{0,0,2,2,2,112,113,0,2},//
{0,0,3,3,3,112,113,0,3},
{0,0,5,5,5,5,5,0,5},
{0,0,6,6,6,6,6,0,6},
{0,0,8,8,8,8,8,0,8}};//19
·GOTO表，二维数组表示：
//A V E T F
int GOTO[20][5]={{1,2,0,0,0},
{0,0,0,0,0},//1
{0,0,0,0,0},
{0,0,0,0,0},
{0,0,5,6,7},
{0,0,0,0,0},//5
{0,0,0,0,0},
{0,0,0,0,0},
{0,0,14,6,7},
{0,0,0,15,7},
{0,0,0,16,7},//10
{0,0,0,0,17},
{0,0,0,0,18},
{0,0,0,0,0},
{0,0,0,0,0},
{0,0,0,0,0},//15
{0,0,0,0,0},
{0,0,0,0,0},
{0,0,0,0,0},
{0,0,0,0,0}};//19
规约时所用到的函数，分别对应每一条规则：
void R1(); //A→V=E
void R2(); //E→E+T
void R3(); //E→E-T
void R4(); //E→T
void R5(); //T→T*F
void R6(); //T→T/F
void R7(); //T→F
void R8(); //F→(E)
void R9(); //F→i
void R10(); //V→i
void R1() {
sign.pop(); sign.pop(); sign.pop(); //弹出符号栈
state.pop(); state.pop(); state.pop(); //弹出状态栈
sign.push('A'); //符号'A'入栈
st=state.top();
printf("r1\t");
}
void R2() {
sign.pop(); sign.pop(); sign.pop(); //弹出符号栈
state.pop(); state.pop(); state.pop(); //弹出状态栈
sign.push('E'); st=state.top(); //符号'E'入栈
printf("r2\t\t");
}
void R3() {
sign.pop(); sign.pop(); sign.pop();
state.pop(); state.pop(); state.pop();
sign.push('E');st=state.top();
printf("r3\t\t");
}
void R4() {
sign.pop();
state.pop();
sign.push('E');st=state.top();
printf("r4\t\t");
}
void R5() {
sign.pop(); sign.pop(); sign.pop();
state.pop(); state.pop(); state.pop();
sign.push('T');st=state.top();
printf("r5\t\t");
}
void R6() {
sign.pop(); sign.pop(); sign.pop();
state.pop(); state.pop(); state.pop();
sign.push('T');st=state.top();
printf("r6\t\t");
}
void R7() {
sign.pop();
state.pop();
sign.push('T');st=state.top();
printf("r7\t\t");
}
void R8() {
sign.pop(); sign.pop(); sign.pop();
state.pop(); state.pop(); state.pop();
sign.push('F');st=state.top();
printf("r8\t\t");
}
void R9() {
sign.pop();
state.pop();
sign.push('F');st=state.top();
printf("r9\t\t");
}
void R10() {
sign.pop();
state.pop();
sign.push('V');st=state.top();
printf("r10\t\t");
}
SLR1分析处理函数：
void SLR()
{
printf("输入串\t\t状态栈\t\t符号栈\t\tACTION\t\tGOTO ");
int i,j,k=1;
state.push(0); //初始化
sign.push('#');
int which; //对应表项内容
char c; //输入符号串首
int a; //坐标
int b;
do{
printf("\n");
c=m[k-1].temp[0]; //输入符号串首
cout<<c<<' ';
for(int j=k;j<=count;j++)
printf("%s",m[j].temp);
printf("\t\t");
displayStack(state);
displayStack1(sign);
a=state.top(); //坐标
b=isVt(c);
/*if(isOp(c)!=-1)
temp1=c;
place.push(temp1);*/
if(b!=-1) //输入串首c是终结符
{
which=ACTION[a][b];
if(which==-1)
{
printf(" acc，分析成功!\n");
flag=1;
break;
}
else if(which==0)
{ printf("error项1\n ");break; }
else if(which>=100) //移进
{
which=s_r(which);
printf("s%d\t\t",which);
sign.push(c);
state.push(which);
k++;
}
else
{
switch(which) //which整型，case不要加''
{
case 1:R1();break;
case 2:R2();break;
case 3:R3();break;
case 4:R4();break;
case 5:R5();break;
case 6:R6();break;
case 7:R7();break;
case 8:R8();break;
case 9:R9();break;
case 10:R10();break;
default:printf("which=%derror项2\n ");break;
}
//状态转移 Vn
int e=isVn(sign.top());
if(e!=-1)
{
int convert=GOTO[st][e];
state.push(convert);
printf("GOTO[%d,%c]=%d",st,sign.top(),convert);
}
}
}
else
{ printf("error_b ");break; }
}while(which!=-1);//while
}
5、实验测试
1.测试用例：i=(i-i*i)#,输入file.txt直接从文件读取输入串，得到结果如下：
四元式结果输出：
由于图片无法上传便罢
6、 实验总结
本次实验是对理论课上所学知识的应用，重点是理解分析栈和符号栈，这里我采用自行造ACTION和GOTO表，这样SLR分析表就出来了，自动造表还是比较复杂。而且在造表的过程中经常出错，最后在大家的讨论中解决了。造完表后的分析过程并不复杂，按部就班分情况来处理。
本次实验加深了我对SLR1的分析过程的理解，也加深了对四元式的认识。
7、源代码
分为两个CPP
Siyuanshi.cpp
#include "stdafx.h"
#include<stdlib.h>
#include<fstream>
#include<iostream>
#include<stdio.h>
using namespace std;
#define MAX 100
int mm=0,sum=0;//sum用于计算运算符的个数
//m用于标输入表达式中字符的个数
char JG='A';
char str[MAX];//用于存输入表达式
int token=0;//左括号的标志
/***********用于更改计算后数组中的值**************/
void change(int e)
{
int f=e+2;
char ch=str[f];
if(ch>='A'&&ch<='Z')
{
for(int l=0;l<mm+10;l++)
{
if(str[l]==ch)
str[l]=JG;
}
}
if(str[e]>='A'&&str[e]<='Z')
{
for(int i=0;i<mm;i++)
{
if(str[i]==str[e])
str[i]=JG;
}
}
}
void chengchuchuli(int i,int mm)
{
i++;
for( ;i<=mm-1;i++)//处理乘除运算
{
if(str[i]=='*'||str[i]=='/')
{
cout<<"("<<str[i]<<" "<<str[i-1]<<" "<<str[i+1]<<" "<<JG<<")"<<endl;
change(i-1);
str[i-1]=str[i]=str[i+1]=JG;
sum--;
JG=(char)(int)JG++;
}
}
}
void jiajianchuli(int j,int mm)
{
j++;
for( ;j<=mm-1;j++)//处理加减运算
{
if(str[j]=='+'||str[j]=='-')
{
cout<<"("<<str[j]<<" "<<str[j-1]<<" "<<str[j+1]<<" "<<JG<<")"<<endl;
change(j-1);
str[j-1]=str[j]=str[j+1]=JG;
sum--;
JG=(char)(int)JG++;
}
}
}
/*扫描遍从文件中读入表达式*/
void scan(FILE *fin)
{
int p[MAX];
char ch='a';
int c=-1,q=0;
while(ch!=EOF)
{
ch=getc(fin);
while(ch==' '||ch=='\n'||ch=='\t') ch=getc(fin);//消除空格和换行符
str[mm++]=ch;
if(ch=='='||ch=='+'||ch=='-'||ch=='*'||ch=='/') sum++;
else if(ch=='(')
{
p[++c]=mm-1;
}
else if(ch==')')
{
q=mm-1;
chengchuchuli(p[c],q);//从左括号处理到又括号
jiajianchuli(p[c],q);
JG=(char)(int)JG--;
str[p[c]]=str[mm-1]=JG;
c--;
JG=(char)(int)JG++;
}
}
}
void siyuanshi()
{
for(int i=0;i<=mm-1;i++)//处理乘除运算
{
if(str[i]=='*'||str[i]=='/')
{
cout<<"("<<str[i]<<" "<<str[i-1]<<" "<<str[i+1]<<" "<<JG<<")"<<endl;
change(i-1);
str[i-1]=str[i]=str[i+1]=JG;
sum--;
JG=(char)(int)JG++;
}
}
for(int j=0;j<=mm-1;j++)//处理加减运算
{
if(str[j]=='+'||str[j]=='-')
{
cout<<"("<<str[j]<<" "<<str[j-1]<<" "<<str[j+1]<<" "<<JG<<")"<<endl;
change(j-1);
str[j-1]=str[j]=str[j+1]=JG;
sum--;
JG=(char)(int)JG++;
}
}
for(int k=0;k<=mm-1;k++)//处理赋值运算
{
if(str[k]=='=')
{
JG=(char)(int)--JG;
cout<<"("<<str[k]<<" "<<str[k+1]<<" "<<" "<<" "<<str[k-1]<<")"<<endl;
sum--;
change(k+1);
str[k-1]=JG;
}
}
}
extern void MAIN(){
char in[MAX]; //用于接收输入输出文件名
FILE *fin;
cout<<"请输入源文件名(包括后缀名）"<<endl;
// scanf("%s",in);
cin>>in;;
if ((fin=fopen(in,"r"))==NULL)
{
cout<<"error"<<endl;
}
cout<<"*********四元式如下*********"<<endl;
scan(fin);//调用函数从文件中读入表达式
siyuanshi();
if(sum==0) printf("成功?");
else printf("有错误");
//关闭文件
fclose(fin);
system("pause");
}
Bianyi_5.cpp
// bianyi_5.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include <iostream>
#include <string>
#include <stack>
#include <map>
#include <string>
#include <cstring>
#include <iomanip>
using namespace std;
extern void MAIN();
#define ADD 1
#define SUB 2
#define MUL 3
#define FH 4
#define SG 5
#define ID 6
#define INT 7
#define LT 8
#define LE 9
#define EQ 10
#define NE 11
#define GT 12
#define GE 13
#define MHEQ 14
#define XGMUL 15
#define ZKH 16
#define YKH 17
#define DIV 18
#define EQ 19//=
#define blz 00
#define COUNT 40
char* keyword[]={"begin","end","if","then","else","for","do","and","or","not"};//保留字
int count;
struct eryuanzu
{
int a;
char temp[COUNT];
}m[COUNT];
void out(int a,char* temp){// 打印二元组
printf("< %d %s >\n",a,temp);
m[count].a=a;
strcpy(m[count].temp,temp); //
count++;
}
stack<int> state; //状态栈
stack<char> sign; //符号栈
char st; //规约弹出时，状态栈顶元素
int flag=0; //标志是否是SLR
stack<string> place; //变量地址栈
/* i ( ) + - * / = #
以1开头的百位数为s移进项，0为error，-1为accept，其余的一位两位数是r规约项*/
int ACTION[20][9]={{103,0,0,0,0,0,0,0,0},//0
{0,0,0,0,0,0,0,0,-1},
{0,0,0,0,0,0,0,104,0},
{0,0,0,0,0,0,0,10,0},
{109,108,0,0,0,0,0,0},
{0,0,0,110,111,0,0,0,1},//5
{0,0,4,4,4,112,113,0,4},
{0,0,7,7,7,7,7,0,7},
{109,108,0,0,0,0,0,0,0},
{0,0,9,9,9,9,9,0,9},
{109,108,0,0,0,0,0,0,0},//10
{109,108,0,0,0,0,0,0,0},
{109,108,0,0,0,0,0,0,0},
{109,108,0,0,0,0,0,0,0},
{0,0,119,110,111,0,0,0,0},
{0,0,2,2,2,112,113,0,2},//
{0,0,3,3,3,112,113,0,3},
{0,0,5,5,5,5,5,0,5},
{0,0,6,6,6,6,6,0,6},
{0,0,8,8,8,8,8,0,8}};//19
//A V E T F
int GOTO[20][5]={{1,2,0,0,0},
{0,0,0,0,0},//1
{0,0,0,0,0},
{0,0,0,0,0},
{0,0,5,6,7},
{0,0,0,0,0},//5
{0,0,0,0,0},
{0,0,0,0,0},
{0,0,14,6,7},
{0,0,0,15,7},
{0,0,0,16,7},//10
{0,0,0,0,17},
{0,0,0,0,18},
{0,0,0,0,0},
{0,0,0,0,0},
{0,0,0,0,0},//15
{0,0,0,0,0},
{0,0,0,0,0},
{0,0,0,0,0},
{0,0,0,0,0}};//19
void R1(); //A→V=E
void R2(); //E→E+T
void R3(); //E→E-T
void R4(); //E→T
void R5(); //T→T*F
void R6(); //T→T/F
void R7(); //T→F
void R8(); //F→(E)
void R9(); //F→i
void R10(); //V→i
int isOp(char a) //判断二元运算符及二元运算符的优先级
{
int i;
switch(a)
{
case '=':i=0;break;
case '+':i=1;break;
case '-':i=1;break;
case '*':i=2;break;
case '/':i=2;break;
default:i=-1;break;
}
return i;
}
int isVt(char a)
{
int i;
switch(a)
{
case 'i':i=0;break;
case '(':i=1;break;
case ')':i=2;break;
case '+':i=3;break;
case '-':i=4;break;
case '*':i=5;break;
case '/':i=6;break;
case '=':i=7;break;
case '#':i=8;break;
default:i=-1;break;
}
return i;
}
int isVn(char a)
{
int i;
switch(a)
{
case 'A':i=0;break;
case 'V':i=1;break;
case 'E':i=2;break;
case 'T':i=3;break;
case 'F':i=4;break;
default:i=-1;break;
}
return i;
}
int s_r(int i) //移进或者其他
{
int result;
if(i/100==1) //移进
result=i-100;
else
result=i;
return result;
}
bool invertStack(stack<int> &one_stack)
{
if (one_stack.empty())//if the stack is null,then don't invert it
{
return false;
}
else
{
//init a stack to save the inverted stack
stack<int> invert;
while (!one_stack.empty())
{
invert.push(one_stack.top());
one_stack.pop();
}
//this moment the stack's inverted state is the stack invert ,so get it back
one_stack = invert;
return true;
}
}
void displayStack(stack<int> one_stack) //打印输出
{
invertStack(one_stack);
while (!one_stack.empty())
{
cout << one_stack.top();
one_stack.pop();
}
cout << '\t' << '\t' ;
}
bool invertStack1(stack<char> &one_stack)
{
if (one_stack.empty())//if the stack is null,then don't invert it
{
return false;
}
else
{
//init a stack to save the inverted stack
stack<char> invert;
while (!one_stack.empty())
{
invert.push(one_stack.top());
one_stack.pop();
}
//this moment the stack's inverted state is the stack invert ,so get it back
one_stack = invert;
return true;
}
}
void displayStack1(stack<char> one_stack)
{
invertStack1(one_stack);
while (!one_stack.empty())
{
cout << one_stack.top();
one_stack.pop();
}
cout << '\t' << '\t';
}
void SLR()
{
printf("输入串\t\t状态栈\t\t符号栈\t\tACTION\t\tGOTO ");
int i,j,k=1;
state.push(0); //初始化
sign.push('#');
int which; //对应表项内容
char c; //输入符号串首
int a; //坐标
int b;
do{
printf("\n");
c=m[k-1].temp[0]; //输入符号串首
cout<<c<<' ';
for(int j=k;j<=count;j++)
printf("%s",m[j].temp);
printf("\t\t");
displayStack(state);
displayStack1(sign);
a=state.top(); //坐标
b=isVt(c);
/*if(isOp(c)!=-1)
temp1=c;
place.push(temp1);*/
if(b!=-1) //输入串首c是终结符
{
which=ACTION[a][b];
if(which==-1)
{
printf(" acc，分析成功!\n");
flag=1;
break;
}
else if(which==0)
{ printf("error项1\n ");break; }
else if(which>=100) //移进
{
which=s_r(which);
printf("s%d\t\t",which);
sign.push(c);
state.push(which);
k++;
}
else
{
switch(which) //which整型，case不要加''
{
case 1:R1();break;
case 2:R2();break;
case 3:R3();break;
case 4:R4();break;
case 5:R5();break;
case 6:R6();break;
case 7:R7();break;
case 8:R8();break;
case 9:R9();break;
case 10:R10();break;
default:printf("which=%derror项2\n ");break;
}
//状态转移 Vn
int e=isVn(sign.top());
if(e!=-1)
{
int convert=GOTO[st][e];
state.push(convert);
printf("GOTO[%d,%c]=%d",st,sign.top(),convert);
}
}
}
else
{ printf("error_b ");break; }
}while(which!=-1);//while
}
void R1() {
sign.pop(); sign.pop(); sign.pop(); //弹出符号栈
state.pop(); state.pop(); state.pop(); //弹出状态栈
sign.push('A'); //符号'A'入栈
st=state.top();
printf("r1\t");
}
void R2() {
sign.pop(); sign.pop(); sign.pop(); //弹出符号栈
state.pop(); state.pop(); state.pop(); //弹出状态栈
sign.push('E'); st=state.top(); //符号'E'入栈
printf("r2\t\t");
}
void R3() {
sign.pop(); sign.pop(); sign.pop();
state.pop(); state.pop(); state.pop();
sign.push('E');st=state.top();
printf("r3\t\t");
}
void R4() {
sign.pop();
state.pop();
sign.push('E');st=state.top();
printf("r4\t\t");
}
void R5() {
sign.pop(); sign.pop(); sign.pop();
state.pop(); state.pop(); state.pop();
sign.push('T');st=state.top();
printf("r5\t\t");
}
void R6() {
sign.pop(); sign.pop(); sign.pop();
state.pop(); state.pop(); state.pop();
sign.push('T');st=state.top();
printf("r6\t\t");
}
void R7() {
sign.pop();
state.pop();
sign.push('T');st=state.top();
printf("r7\t\t");
}
void R8() {
sign.pop(); sign.pop(); sign.pop();
state.pop(); state.pop(); state.pop();
sign.push('F');st=state.top();
printf("r8\t\t");
}
void R9() {
sign.pop();
state.pop();
sign.push('F');st=state.top();
printf("r9\t\t");
}
void R10() {
sign.pop();
state.pop();
sign.push('V');st=state.top();
printf("r10\t\t");
}
///////////////////////////////////////////////
void scanner(FILE *p)
{
char filein[40],fileout[40]; //文件名
printf("请输入要打开的源文件名（包括路径）\n");
scanf("%s",filein);
//printf("请输入要输出的目标文件名（包括路径）\n");
//scanf("%s",fileout);
if((p=fopen(filein,"r"))==NULL) {printf("输入文件打开有错！\n");return;}
// if((q=fopen("fileout","rw"))==NULL) {printf("输出文件打开有错！\n");return;}
char token[COUNT]; //输出数组
int r=0,i=0;
char ch;
ch=fgetc(p);
while(!feof(p)) //没有读到文件末尾
{
if(isdigit(ch))
{
i=0;
token[i]=ch;
while(isdigit(ch=fgetc(p)))
{
i++;
token[i]=ch;
}
token[i+1]='\0'; //整数结束。不要忘结束标志!
fseek(p,-1,1); //重定向到当前位置的前一个!
out(INT,token);
//fprintf(q,"%d %s\n",INT,token);
}
else if(isalpha(ch))
{
i=0;
int flag=0; //标志是否是保留字，默认不是
token[i]=tolower(ch);
while(isalpha(ch=fgetc(p)))
{
i++;
token[i]=tolower(ch);
}
token[i+1]='\0';
fseek(p,-1,1);
for(r=0;r<8;r++)
{
if(!strcmp(token,keyword[r]))
{
printf("<blz %s>\n",token);
// fprintf(q,"%d %s\n","保留字",token);
flag=1;
break;
}
}
if(!flag)
{ out(ID,token);
// fprintf(q,"%d %s\n",ID,token);
}
}
else
{
i=0;
switch(ch)
{
case '+':{ token[i]=ch;
token[i+1]='\0';
out(ADD,token);
}break;
case '-':{ token[i]=ch;
token[i+1]='\0';
out(SUB,token);}break;
case '*':{ token[i]=ch;
token[i+1]='\0';
out(MUL,token);}break;
case ';':{ token[i]=ch;
token[i+1]='\0';
out(FH,token);}break;
case '|':{ token[i]=ch;
token[i+1]='\0';
out(SG,token);}break;
case '(':{ token[i]=ch;
token[i+1]='\0';
out(ZKH,token);}break;
case ')':{ token[i]=ch;
token[i+1]='\0';
out(YKH,token);}break;
case '=':{ token[i]=ch;
token[i+1]='\0';
out(EQ,token);}break;
case ' ':{ break;} //空格直接跳
case '#':{ break;} //井号用作结尾
case '<':{token[i]=ch;
ch=fgetc(p);
if(ch=='='){
token[i+1]='=';
token[i+2]='\0';
out(LE,token);
}
else if(ch=='>'){
token[i+1]='>';
token[i+2]='\0';
out(NE,token);
}
else
{ printf(" error \n");
fseek(p,-1,1); //多读的要回退一个字符
}
}break;
case '>':{token[i]=ch;
ch=fgetc(p);
if(ch=='=')
{
token[i+1]='=';
token[i+2]='\0';
out(GE,token);
}
else
{ printf(" error \n");
fseek(p,-1,1); //多读的要回退一个字符
}
}break;
case ':':{token[i]=ch;
ch=fgetc(p);
if(ch=='='){
token[i+1]='=';
token[i+2]='\0';
out(MHEQ,token);
}
else
{ printf(" error \n");
fseek(p,-1,1); //多读的要回退一个字符
}
}break;
case '/':{token[i]=ch;
ch=fgetc(p);
if(ch=='*')
{
token[i+1]='*';
token[i+2]='\0';
out(XGMUL,token);
while(1) //注释部分的处理！
{
ch=fgetc(p);
if(ch=='*')
{
if((ch=fgetc(p))=='/')
break;
}
}
}
else // 除号，修改第一次程序部分
{ /*printf(" error \n");
fseek(p,-1,1); */
out(DIV,token);
fseek(p,-1,1);//多读的要回退一个字符
}
}break;
default:{
printf(" error\n ");
}break;
}
}
ch=fgetc(p);
}
fclose(p);
}
int main(int argc, char* argv[])
{
printf("编译原理实验_5：SLR分析程序（待分析内容在文件file.txt中，以#结尾）\n");
FILE *fin,*q;
scanner(fin);
strcpy(m[count].temp,"#");//!
count=count+1;
//scanner(p,q);
SLR();
MAIN();
return 0;
}
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