SLR分析器的设计
作者：互联网
一、实验目的
根据文法编制SLR语法分析程序，以便对输入的符号串进行语法分析。通过编写SLR语法分析程序掌握移进归约方法的基本原理、SLR分析表的构造方法以及移进归约分析法主控程序的设计。
二、实验内容
对下列算术表达式的文法编写SLR语法分析程序，要求对输入的符号串进行语法分析：
（1）E->E+T
（2）E->E-T
（3）E->T
（4）T->T*F
（5）T->T/F
（6）T->F
（7）F->(E)
（8）F->id
（9）F->num
注：文法中id表示标识符（此处标识符的定义与实验一中标识符的定义相同），num表示数字（简单处理可以认为是整数）
三、实验设计方案
五、测试方案及测试结果
结语
SLR分析器设计的介绍就到这里啦，希望这篇文章能给予你一些帮助，感谢各位人才的：点赞、收藏和评论，我们下次见。
附录
以下提供测试代码
SLR语法分析器
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <windows.h>
char Keywords[22][10]=
{//关键字表
"double", "int", "struct", "break", "static",
"long", "switch", "case", "char", "return",
"const", "float", "short", "continue", "for",
"signed", "void","default", "do", "while",
"if", "else"};
char Symbol[30][10] =
{//符号表
"+", "-", "*", "/", "=", "==", "!=", ">", "<", ">=",
"<=", ",", ";", "{", "}", "(", ")", "!","&", "&&",
"|", "||", "^", "#", "%", "[", "]", ".", ":", "$"};
static char word[20];//每次扫描提取到的词
bool IsLetter(char ch)
{//判断是否为字母，包括下划线
if (ch >= 'a'&&ch <= 'z' || ch >= 'A'&&ch <= 'Z'|| ch=='_')
return true;
else
return false;
}
bool IsDigit(char ch)
{//判断是否为数字
if (ch >= '0' && ch <= '9')
return true;
else
return false;
}
bool IsSymbol(char ch)
{//判断是否为单字符号
if (ch == '+' || ch == '-' || ch == '*' || ch == '/' || ch == ';'
|| ch == '(' || ch == ')' || ch == '^' || ch == ',' || ch == '#'
|| ch == '%' || ch == '[' || ch == ']' || ch == '{' || ch == '}'
|| ch == '.' || ch == ':'|| ch == '$')
return true;
else
return false;
}
int IsKeyword(char word[])
{//判断是否为关键字
for(int i=0;i<32;i++)
if (strcmp(word,Keywords[i])==0)
return i + 1;
return -1;
}
int LexicalAnalyzer(char CurrentCode[], int &cc)
{//词法分析
int i, w = 0, cn;
while (CurrentCode[cc] == ' ')
cc++;
for (i = 0; i<20; i++)
word[i] = '\0';
if (IsLetter(CurrentCode[cc]))
{
word[w] = CurrentCode[cc];
w++;
cc++;
while (IsLetter(CurrentCode[cc]) || IsDigit(CurrentCode[cc]))
{
word[w] = CurrentCode[cc];
w++;
cc++;
}
word[w] = '\0';
cn = IsKeyword(word);
if(cn == -1)
return 33;
else
return cn;
}
else if (IsDigit(CurrentCode[cc]))
{
while (IsDigit(CurrentCode[cc]))
{
word[w] = CurrentCode[cc];
w++;
cc++;
}
word[w] = '\0';
return 34;
}
else if (IsSymbol(CurrentCode[cc]))
{
word[0] = CurrentCode[cc];
word[1] = '\0';
cc++;
for (i = 0; i<30; i++)
if (strcmp(word, Symbol[i]) == 0)
return 35 + i;
}
else if (CurrentCode[cc] == '<')
{
cc++;
if (CurrentCode[cc] == '=')
{
cc++;
return 45;
}
else
return 43;
}
else if (CurrentCode[cc] == '>')
{
cc++;
if (CurrentCode[cc] == '=')
{
cc++;
return 44;
}
else
return 42;
}
else if (CurrentCode[cc] == '=')
{
cc++;
if (CurrentCode[cc] == '=')
{
cc++;
return 40;
}
else
return 39;
}
else if (CurrentCode[cc] == '!')
{
cc++;
if (CurrentCode[cc] == '=')
{
cc++;
return 41;
}
else
return 52;
}
else if (CurrentCode[cc] == '&')
{
cc++;
if (CurrentCode[cc] == '&')
{
cc++;
return 54;
}
else
return 53;
}
else if (CurrentCode[cc] == '|')
{
cc++;
if (CurrentCode[cc] == '|')
{
cc++;
return 56;
}
else
return 55;
}
else if (CurrentCode[cc] == '\0')
{
cc++;
return 0;
}
else
{
cc++;
return -2;
}
}
typedef struct
{//栈的结构体
char *base;
char *top;
int stacksize;
}Stack;
int InitStack(Stack &S)
{//创造一个空栈
S.base=(char *)malloc(100*sizeof(char));
if(!S.base)
exit(-2);
S.top=S.base;
S.stacksize=100;
return 1;
};
int DestroyStack(Stack &S)
{//销毁栈S
while(S.top!=S.base)
free(--S.top);
return 1;
}
int StackEmpty(Stack S)
{//判断是否为空栈
if(S.top==S.base)
return 1;
else
return 0;
}
int GetTop(Stack S,char &e)
{//若栈不为空，取栈顶元素e
if(S.base==S.top)
return 0;
e=*(S.top-1);
return 1;
}
int Push(Stack &S,char e)
{//将e压入栈顶
*S.top=e;
S.top++;
return 1;
}
int Pop(Stack &S,char &e)
{//若栈不为空，将栈顶元素e出栈
if(S.top==S.base)
return 0;
e=*--S.top;
return 1;
}
int StackTraverse(Stack S)
{//遍历栈中的元素
if(S.base==NULL)
return -1;
if(S.base==S.top)
printf("栈中没有元素\n");
char *p;
p=S.top;
while(p-S.base>0)
{
p--;
printf("%c ",*p);
}
printf("\n");
return 1;
}
int ReverseTraverse(Stack S)
{//遍历输出从栈底至栈顶的元素
if(S.base==NULL)
return -1;
if(S.base==S.top)
printf("\t栈中没有元素\n");
char *p,a[30];
int i=0;
p=S.base;
while(p<S.top)
{
a[i]=*p;
i++;
p++;
}
a[i]='\0';
printf("\t%-15s",a);
return 1;
}
void SetFontRed()
{//字体变红
SetConsoleTextAttribute(GetStdHandle(STD_OUTPUT_HANDLE), FOREGROUND_INTENSITY | FOREGROUND_RED);
}
void SetFontWhite()
{//字体变白
SetConsoleTextAttribute(GetStdHandle(STD_OUTPUT_HANDLE), FOREGROUND_INTENSITY | FOREGROUND_RED | FOREGROUND_GREEN | FOREGROUND_BLUE);
}
char row[17]=
{//SLR分析表中的行头
'0','1','2','3','4','5','6','7','8','9','a','b','c','d','e','f','g'
};
char column[12]=
{//SLR分析表中的列头
'i','n','+','-','*','/','(',')','$','E','T','F'
};
char TableValue[10];//读取到的SLR分析表中的内容
char SLR1Table[17][12][10]=
{ //将SLR分析表输入进行存储
"s5","s6","e1","e1","e1","e1","s4","e2","e1","1","2","3",
"e3","e3","s7","s8","0","0","e3","e2","acc","0","0","0",
"e3","e3","r3","r3","s9","sa","e3","r3","r3","0","0","0",
"e3","e3","r6","r6","r6","r6","e3","r6","r6","0","0","0",
"s5","s6","e1","e1","e1","e1","s4","e1","e1","b","2","3",
"0","0","r8","r8","r8","r8","e3","r8","r8","0","0","0",
"e3","0","r9","r9","r9","r9","e3","r9","r9","0","0","0",
"s5","s6","e1","e1","e1","e1","s4","e1","e1","0","c","3",
"s5","s6","e1","e1","e1","e1","s4","e1","e1","0","d","3",
"s5","s6","e1","e1","e1","e1","s4","e1","e1","0","0","e",
"s5","s6","e1","e1","e1","e1","s4","e1","e1","0","0","f",
"e3","e3","s7","s8","0","0","e3","sg","e4","0","0","0",
"e3","e3","r1","r1","s9","sa","e3","r1","r1","0","0","0",
"e3","e3","r2","r2","s9","sa","e3","r2","r2","0","0","0",
"e3","e3","r4","r4","r4","r4","e3","r4","r4","0","0","0",
"e3","e3","r5","r5","r5","r5","e3","r5","r5","0","0","0",
"r7","r7","r7","r7","r7","r7","r7","r7","r7","0","0","0"};
char Products[10][10]=
{//按顺序存储表达式
"S->E","E->E+T","E->E-T","E->T","T->T*F","T->T/F","T->F","F->(E)","F->i","F->n"
};
int ElementSearch(char a[],char e,int length)
{//查找表头，返回其对应的下标
int i,result=-1;
for(i=0;i<length;i++)
{
if(a[i]==e)
{
result=i;
break;
}
}
return result;
}
int TableSearch(char SLR1Table[17][12][10],char e,char element,int ROW,int COLUMN)
{//从SLR分析表中查找
int RowIndex=ElementSearch(row,e,ROW);
int ColIndex=ElementSearch(column,element,COLUMN);
if(RowIndex==-1 || ColIndex==-1)
return -1;//异常
else
{
if(strcmp(SLR1Table[RowIndex][ColIndex],"0")==0)//空，出错
return 0;
else if (SLR1Table[RowIndex][ColIndex][0]=='s')//移进
{
strcpy(TableValue,SLR1Table[RowIndex][ColIndex]);
return 1;
}
else if (SLR1Table[RowIndex][ColIndex][0]=='r')//归约
{
strcpy(TableValue,SLR1Table[RowIndex][ColIndex]);
return 2;
}
else if(strcmp(SLR1Table[RowIndex][ColIndex],"acc")==0)//接受
return 4;
else if (SLR1Table[RowIndex][ColIndex][0]=='e')//错误
{
strcpy(TableValue,SLR1Table[RowIndex][ColIndex]);
return 5;
}
else
{
strcpy(TableValue,SLR1Table[RowIndex][ColIndex]);
return 3;
}
}
}
int Parser(Stack s,char str[],char SLR1Table[17][12][10],int ROW,int COLUMN)
{//语法分析总控程序
Push(s,'0');
int i=0,j,isTrue=1,result;
char e='0';
SetFontWhite();
printf("\tSLR(1)分析法分析过程：\n\n");
printf("\t分析栈 输入串 动作\n");
while(1)
{
SetFontWhite();
ReverseTraverse(s);
printf("\t%-14s",str+i);
GetTop(s,e);
result=TableSearch(SLR1Table,e,str[i],ROW,COLUMN);
switch(result)
{
case -1:
SetFontRed();
printf("找不到表头\n");
return 0;
case 0:
SetFontRed();
printf("表内为空\n\n");
return 0;
case 1:
SetFontWhite();
Push(s,str[i]); //移进
Push(s,TableValue[1]);
printf("%s\n",TableValue);
i++;
break;
case 2:
for(j=0;j<2*(strlen(Products[TableValue[1]-'0'])-3);j++) //归约
Pop(s,e);
GetTop(s,e);
Push(s,Products[TableValue[1]-'0'][0]);
printf("用%s归约\n",Products[TableValue[1]-'0']);
TableSearch(SLR1Table,e,Products[TableValue[1]-'0'][0],ROW,COLUMN);
Push(s,TableValue[0]);
break;
case 4:
printf("结束\n\n");
return isTrue;
case 5:
SetFontRed();
if(TableValue[1] == '1')//e1：期望输入符号为运算对象，将分析中的产生式归约成 T
{
printf("error：缺少运算对象\n");
while(str[i]!='$'&&str[i]!='+'&&str[i]!='-'&&str[i]!='*'&&str[i]!='/'&&str[i]!=')')
i++;
GetTop(s,e);
while(e!='0'&&e!='4'&&e!='7'&&e!='8')
{
Pop(s,e);
Pop(s,e);
GetTop(s,e);
}
if(e == '0' ||e == '4')
{
Push(s,'T');
Push(s,'2');
}
else if(e == '7')
{
Push(s,'T');
Push(s,'c');
}
else if(e == '8')
{
Push(s,'T');
Push(s,'d');
}
}
else if(TableValue[1] == '2')//e2：遇到不配对的右括号，将括号内容归约成 T
{
printf("error：不配对的右括号\n");
GetTop(s,e);
while(e!='0'&&e!='4'&&e!='7'&&e!='8')
{
Pop(s,e);
Pop(s,e);
GetTop(s,e);
}
if(e == '0' ||e == '4')
{
Push(s,'T');
Push(s,'2');
}
else if(e == '7')
{
Push(s,'T');
Push(s,'c');
}
else if(e == '8')
{
Push(s,'T');
Push(s,'d');
}
i++;
}
else if(TableValue[1] == '3')//e3：期望运算符，将分析中的产生式归约成 T
{
printf("error：缺少运算符\n");
while(str[i]!='$'&&str[i]!='+'&&str[i]!='-'&&str[i]!='*'&&str[i]!='/'&&str[i]!=')')
i++;
GetTop(s,e);
while(e!='0'&&e!='4'&&e!='7'&&e!='8')
{
Pop(s,e);
Pop(s,e);
GetTop(s,e);
}
if(e == '0' ||e == '4')
{
Push(s,'T');
Push(s,'2');
}
else if(e == '7')
{
Push(s,'T');
Push(s,'c');
}
else if(e == '8')
{
Push(s,'T');
Push(s,'d');
}
}
else if(TableValue[1] == '4')//e4：根据情况归约成T或F
{
printf("error：缺少右括号\n");
GetTop(s,e);
while(e!='0'&&e!='4'&&e!='7'&&e!='8'&&e!='9'&&e!='a')
{
Pop(s,e);
Pop(s,e);
GetTop(s,e);
}
if(e == '0'||e == '7'||e == '8')
{
Push(s,'F');
Push(s,'3');
}
else if(e == '4')
{
Pop(s,e);
Pop(s,e);
GetTop(s,e);
if(e == '9'||e == 'a')
{
Pop(s,e);
Pop(s,e);
}
else
{
Push(s,'F');
Push(s,'3');
}
}
else if(e == '9')
{
Push(s,'F');
Push(s,'e');
}
else if(e == 'a')
{
Push(s,'F');
Push(s,'f');
}
}
isTrue = 0;
}
}
}
int main()
{
int ROW=17,COLUMN=12;
int i=0,si=0,lex,result;
char str[30],newstr[30];
printf("请输入字符串(格式为字符串$)：");
gets(str);
printf("\n");
while(str[si]!='\0')
{
lex = LexicalAnalyzer(str, si);
if(lex == 33)
newstr[i]='i';
else if(lex == 34)
newstr[i]='n';
else if(lex == 64)
newstr[i]='$';
else
newstr[i]=str[i];
i++;
}
newstr[i]='\0';
Stack s;
InitStack(s);
result=Parser(s,newstr,SLR1Table,ROW,COLUMN);
SetFontWhite();
if(result==0)
printf("\t由上表可知，%s是错误的表达式!\n",str);
else
printf("\t由上表可知，%s是正确的表达式!\n",str);
return 0;
}
标签：return,int,分析器,SLR,else,Push,设计,e1,e3 来源： https://blog.csdn.net/qq_44075108/article/details/121949777