语法分析实验 语法分析(LR)

实验报告

课程名称编译原理

实验名称语法分析(LR)

班级计科1203

学号201207010321

姓名马万里

指导教师黄全舟

实验日期2015.5.21 二、实验原理

三、主要仪器设备

Pc机一台(装有vs2012)

四、主要代码及其说明

#include

#include

#include

using namespace std;

#define N 10

char*Action[][4]=

{//此处三维也可以

{"S3","S4","ERROR","ERROR"},

{"ERROR","ERROR","ERROR","acc"},

{"ERROR","ERROR","S5","R2"},

{"ERROR","ERROR","R3","R3"},

{"ERROR","ERROR","R4","R4"},

{"S3","S4","ERROR","ERROR"},

{"ERROR","ERROR","ERROR","R1"}};

Int GoTo[][2]={{1,2},{0,0},{0,0},{0,0},{0,0},{6,2},{0,0}};

char*GL[]={"L->E,L","L->E","E->a","E->b"};

int_Push(vector&State,vector&InChar,int a,char b) {

State.push_back(a);

InChar.push_back(b);

return a;

}

void_Pop(vector&State,vector&InChar,int cout)

{

while(cout)

{

State.pop_back();

InChar.pop_back();

cout--;

}

}

void PrinStack(vectorState,vectorInChar)//状态栈和栈中符号输出

{

vector::iterator iter_s=State.begin();

vector::iterator iter_i=InChar.begin();

for(;iter_s!=State.end();iter_s++)

{

cout<<*iter_s;

}

printf("\t");

for(;iter_i!=InChar.end();iter_i++)

{

cout<<*iter_i;

}

printf("\t\t");

}

void printInCh(char*InCh)//余留符号输出

{

while(*InCh!='\0')

{

printf("%c",*InCh);

InCh++;

}

printf("\t\t");

}

void printAanly(char a,int b)//输出分析动作

{

if(a=='S')

printf("S%d",b);

if(a=='R')

printf("R%d",b);

printf("\t\t");

}

void printNextState(char a,const char b)//输出下一状态

{

if(a=='S')

printf("%d",atoi(&b));

}

int Lcount(char*p)//判断文法右边符号的个数

{

int m=0;

while(*p!='\0')

{

if(*p=='>') m=1;

if(m>0) m++;

p++;

}

return(m-2);

}

void menu()

{

printf("步骤\t状态\t栈中符号\t余留符号\t分析动作\t下一状态"); }

void main()

{

cout<<"请输入符号串(\"以#标记结束\"):";

vectorState;//状态栈

vectorInChar;//栈中的符号

State.push_back(0);

InChar.push_back('#');

char*InCh;//输入的字符串InCh=(char*)malloc(sizeof(char)*N);

gets(InCh);

printf("\n****************************符号串%s的LR分析过程****************************\n\n",InCh);

menu();

int ToS=0,InS=0,count=0;

while(strncmp(Action[ToS][InS],"acc",3))

{

count++;//步骤序号

if(!strncmp(Action[ToS][InS],"ERROR",5))

{

printf("error");

break;

}

else if(!strncmp(Action[ToS][InS],"S",1))

{

int temp;

printf("\n%d\t",count);

PrinStack(State,InChar);

switch(Action[ToS][InS][1])

{

case'1':temp=_Push(State,InChar,1,*InCh);break;

case'2':temp=_Push(State,InChar,2,*InCh);break;

case'3':temp=_Push(State,InChar,3,*InCh);break;

case'4':temp=_Push(State,InChar,4,*InCh);break;

case'5':temp=_Push(State,InChar,5,*InCh);break;

}

printInCh(InCh);

printAanly('S',atoi(&Action[ToS][InS][1]));

printNextState('S',Action[ToS][InS][1]);

ToS=temp;

InCh++;

if(*InCh=='a')InS=0;

if(*InCh=='b')InS=1;

if(*InCh==',')InS=2;

if(*InCh=='#')InS=3;

}

else if(!strncmp(Action[ToS][InS],"R",1))

{

int itemp;

int ntemp;

int temp;

char ctemp;

printf("\n%d\t",count);

PrinStack(State,InChar);

int f=atoi(&Action[ToS][InS][1])-1;

itemp=Lcount(GL[f]);//右部的符号个数

_Pop(State,InChar,itemp);//删除两栈中的tem个字符

ctemp=*GL[f];//获取当前文法的左边字符

if(ctemp=='L')

temp=0;

if(ctemp=='E')

temp=1;

ToS=State.back();

ntemp=GoTo[ToS][temp];//获取当前的下一状态

_Push(State,InChar,ntemp,ctemp);

printInCh(InCh);

printAanly('R',f+1);

printf("GOTO[%d,%c]=%d",ToS,ctemp,ntemp);

ToS=ntemp;

}

}

count++;

//处理当acc时的情况

if(!strncmp(Action[ToS][InS],"acc",3))

{

printf("\n%d\t",count);

PrinStack(State,InChar);

printInCh(InCh);

printf("acc\n");

}

五、程序测试结果

输入:a,b,a#的执行过程如下

输入:a,,#的执行过程如下

本次试验相对来说比较有挑战性,但是如果利用现成的vs2012中动态数组(其底层实质还是栈)来实现还是比较轻松的,而且这只是语言语法方面,重要的是理解LR分析法的过程及其思想.而且在逻辑方面也要很清楚.总的来说这次试

验比较成功.