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标题:关于 使用词法分析,语法分析将布尔表达式转换为逆波兰式 的下了个程序看不 ...
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陈子风
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关于 使用词法分析,语法分析将布尔表达式转换为逆波兰式 的下了个程序看不太懂
#include<ctype.h>
#include<stdlib.h>
#include<iomanip>
#include<string>
#include<iostream.h>
struct danci
  {
      int id;            //单词种别
      char name[10];    //单词属性值
  };
int fout[20];
int index=1;
int G_table[9][6];
int k[20];
int n=0;
danci word[20];
int boo(int);
int compare(int);
int not(int);
int forid(int);
typedef struct node{
    int data;
       node* next;
 }* link;
link insert(link stack,int n)//向前插入链表
{
    link newnode;
    newnode=new node;
    if(!newnode)
    {
        cout<<"error!"<<endl;
        exit(0);
    }
    newnode->data=n;
    newnode->next=stack;
    stack=newnode;
    return stack;
}
link del(link stack,int &value)//删除链表元素
{
    link top;
    if(stack)
    {
        top=stack;
        stack=stack->next;
        value=top->data;
        delete top;
        return stack;
    }
    value=-1;
    return NULL;
}
void error()
{
    cout<<"词法错误,标志符不能以数字开头"<<endl;
    exit(0);
}
void error1()
{
    cout<<"语法错误,输入正确句子"<<endl;
    exit(0);
}
int not(int i)
{
    int ii;
    i=forid(i);
    while(word[i].id==3)
    {
        ii=i;
        i++;
        i=forid(i);
        fout[index]=ii;
        index++;
    }
    return i;
}
void init()//初始化文法矩阵
{
    for(int i=0;i<9;i++)                   
    for(int j=0;j<6;j++)
        G_table[i][j]=-1;
    G_table[0][0]=0;G_table[0][1]=1; //初始化文法表
    G_table[1][0]=1;G_table[1][1]=1;G_table[1][2]=5;G_table[1][3]=2;
    G_table[2][0]=1;G_table[2][1]=2;
    G_table[3][0]=1;G_table[3][1]=1;G_table[3][2]=3;
    G_table[4][0]=2;G_table[4][1]=3;
    G_table[5][0]=3;G_table[5][1]=6;G_table[5][2]=3;
    G_table[6][0]=3;G_table[6][1]=9;G_table[6][2]=1;G_table[6][3]=10;
    G_table[7][0]=3;G_table[7][1]=7;
    G_table[8][0]=3;G_table[8][1]=9;G_table[8][2]=7;G_table[8][3]=8;
    G_table[8][4]=7;G_table[8][5]=10;
}
void print()//打印文法
{   
char a0[]="E";char a1[]="S";
char a2[]="B";char a3[]="C";
char a4[]="and";char a5[]="or";
char a6[]="not";char a7[]="id";
char a8[]="relop";char a9[]="(";
char a10[]=")";
cout<<"本实验的文法为:"<<endl;
cout<<a0<<"->"<<a1<<endl;
cout<<a1<<"->"<<a1<<' '<<a5<<a2<<endl;
cout<<a1<<"->"<<a2<<endl;
cout<<a2<<"->"<<a2<<' '<<a4<<' '<<a3<<endl;
cout<<a2<<"->"<<a3<<endl;
cout<<a3<<"->"<<a6<<' '<<a3<<endl;
cout<<a3<<"->"<<a9<<a1<<a10<<endl;
cout<<a3<<"->"<<a7<<endl;
cout<<a3<<"->"<<a9<<a7<<' '<<a8<<' '<<a7<<a10<<endl;
}
int  lex(char * buf)//词法分析
{
    int i=0,j=0;
    int n=0;
    char Temp_name[10];
    while(buf[i]!='\0')
    {
        if(buf[i]==' '||buf[i]=='\n')
            i++;
        else if(isalpha(buf[i])||buf[i]=='_')//标志符判断
        {
            Temp_name[j]=buf[i];
            i++;
            while(isalnum(buf[i]))
            {
                j++;
                Temp_name[j]=buf[i];
                i++;
            }
           // if
            memcpy(word[n].name,Temp_name,j+1);
            word[n].name[j+1]='\0';
            if(strcmp(word[n].name,"and")==0)
                word[n].id=1;
            else if(strcmp(word[n].name,"or")==0)
                word[n].id=2;
            else if(strcmp(word[n].name,"not")==0)
                word[n].id=3;
            else
                word[n].id=0;
            n++;
            j=0;
        }
        else if(isdigit(buf[i]))//数字判断
        {
            while(isdigit(buf[i]))
            {
            Temp_name[j]=buf[i];
            i++;
            j++;
            }
                if(isalpha(buf[i]))
                  error();
            memcpy(word[n].name,Temp_name,j+1);
            word[n].name[j]='\0';
            n++;
            j=0;
        }
        else//运算符号判断
        {
            switch(buf[i])
            {
                case '>':if(buf[i+1]=='=')
                         {
                              i++;
                                 word[n].id=8;
                              strcpy(word[n].name,">=");
                         }
                         else
                         {
                             word[n].id=4;
                             strcpy(word[n].name,">");
                         }
                         n++;break;
                case '<':if(buf[i+1]=='=')
                         {
                              i++;
                                 word[n].id=9;
                              strcpy(word[n].name,"<=");
                         }
                         else
                         {
                             word[n].id=5;
                             strcpy(word[n].name,"<");
                         }
                         n++;break;
                case '=': word[n].id=6;strcpy(word[n].name,"=");n++;break;
                case '!': if(buf[i+1]=='=')
                          {
                              i++;
                                 word[n].id=7;strcpy(word[n].name,"!=");
                              n++;
                          }
                           break;
                case '(': word[n].id=10;strcpy(word[n].name,"(");n++;break;
                case ')': word[n].id=11;strcpy(word[n].name,")");n++;break;
            }
            i++;
        }
    }
    strcpy(word[n].name,"#");
        word[n].id=999;
    return n;//单词符号个数
}

void gram(int * kind)
{
    int n=0;
    link state;                        
    link lex;                     
    link operate;                        
    link ID;                        
    state=new node[30];lex=new node[15];
    if(!state||!lex)               
    {
        cout<<"over stack! "<<endl;
        exit(0);
    }
    int trace=0;                    
    int G[9][5];                    
    int Stable[16][11];            //SLR分析表
    int i,j;
    for(i=0;i<9;i++)                   
        for(j=0;j<5;j++)
            G[i][j]=-1;
    for(i=0;i<18;i++)
        for(j=0;j<11;j++)
            Stable[i][j]=0;        //初始化SLR表,其中表的初值为0,表示出错!                                
    //设置SLR表 1-17表示移进,n进状态栈;21-29表示按N-20个产生式规约;0表示出错
    Stable[0][2]=4;Stable[0][3]=5;Stable[0][6]=6;
    Stable[0][8]=1;Stable[0][9]=2;Stable[0][10]=3;
    Stable[1][0]=7;Stable[1][7]=111;
    Stable[2][0]=23;Stable[2][1]=8;Stable[2][4]=23;Stable[2][7]=23;
    Stable[3][0]=25;Stable[3][1]=25;Stable[3][4]=25;
    Stable[3][7]=25;
    Stable[4][2]=4;Stable[4][3]=5;Stable[4][6]=6;Stable[4][10]=9;
    Stable[5][2]=4;Stable[5][3]=5;Stable[5][6]=11;
    Stable[5][8]=10;Stable[5][9]=2;Stable[5][10]=3;
    Stable[6][0]=28;Stable[6][1]=28;Stable[6][4]=28;Stable[6][7]=28;
    Stable[7][2]=4;Stable[7][3]=5;Stable[7][6]=6;
    Stable[7][9]=12;Stable[7][10]=3;
    Stable[8][2]=4;Stable[8][3]=5;    Stable[8][6]=6;Stable[8][10]=13;
    Stable[9][0]=26;Stable[9][7]=26;Stable[9][4]=26;Stable[9][1]=26;
    Stable[10][0]=7;Stable[10][4]=14;
    Stable[11][0]=28;Stable[11][1]=28;
    Stable[11][4]=28;Stable[11][5]=15;Stable[11][7]=28;
    Stable[12][0]=22;Stable[12][1]=8;Stable[12][4]=22;Stable[12][7]=22;
    Stable[13][1]=4;Stable[13][2]=4;Stable[13][3]=4;
    Stable[13][0]=24;Stable[13][1]=24;Stable[13][4]=24;Stable[13][7]=24;
    Stable[14][0]=27;Stable[14][1]=27;Stable[14][4]=27;Stable[14][7]=27;
    Stable[15][6]=16;Stable[16][6]=17;
    Stable[17][0]=29;Stable[17][1]=29;Stable[17][4]=29;Stable[17][7]=29;
    kind[n]=7;                                //输入串最后字符为#
    n++;
    state=insert(state,0);
    lex=insert(lex,7);
    int a;
    int staTop;                                //存放状态栈的栈顶元素
    int operater;int ch1=0;int ch2;
    for(a=0;a<n;a++)                        //进行SLR分析
    {
        staTop=state->data;
      if(Stable[staTop][kind[a]]==0)
        {

            break;
                
        }
     else if(Stable[staTop][kind[a]]<20)
        {            //规约
            int sta1;
            int letter1;
            if(state!=NULL)
                state=del(state,sta1);
            else
            {
             error1();
            }
            int num=0;                    //出栈次数
            int b;
            for(b=1;b<G_table[Stable[staTop][kind[a]]][4];b++)
            {                            //规约式右部出栈
                lex=del(lex,letter1);
                num++;
            }
            lex=insert(lex,G_table[Stable[staTop][kind[a]]][0]);//规约式左部进栈
            for(b=1;b<num;b++)            //状态栈中再出栈num-1个状态
                state=del(state,sta1);
            int staTop1;
            staTop1=state->data;
            state=insert(state,Stable[staTop1][G_table[Stable[staTop][kind[a]]][0]]-50);
            a--;
            if(num>1)
            {    
                if(ch1==0)//总结符
                {
                    ID=del(ID,ch1);
                    ch1=trace;trace--;
                }
                else
                    ch1=-1;
                operate=del(operate,operater);
                ID=del(ID,ch2);
                ch2=trace;trace--;
                ch1--;
            }
        }
        else if(Stable[staTop][kind[a]]<50)
        {//移进一个符号
            state=insert(state,Stable[staTop][kind[a]]-20);
            lex=insert(lex,kind[a]);
            if(kind[a]==0)
            {
                trace++;
                ID=insert(ID,kind[a]);
            }
            else
                operate=insert(operate,kind[a]);
        }
        else if(Stable[staTop][kind[a]]==99)
        {
            cout<<endl;
            cout<<"ACCEPT!"<<endl;
            break;
        }
    
    }
}
void forboland(int length)
{
    
    int i=0;
    while((i<length)&&(word[i].id!=99))
    {
        i=boo(i);
        if(word[i].id==99)
        {
         error1();
        }
    }

}
int boo (int i)
{
    int ii;
    i=compare(i);
    while (word[i].id==1||word[i].id==2)
    {
        ii=i;
        i++;                
        i=compare(i);    
        fout[index]=ii;//
        index++;
    }
    return i;
}

int forid(int i)
{
    if (word[i].id==0)        
    {
        fout[index]=i;
        index++;
        i++;
    }
    else
    {
        if(word[i].id==10)
        {
            i++;                
            i=boo(i);
            if (word[i].id==11)
                i++;            
            else
                word[i].id=99;
        }
        else if(word[i].id==3)
        {;}
        else
        {
            error1();
            word[i].id=99;
        }
    }
    return i;
}
void boland()
{
    for(int i=1;i<index;i++)
    {
        if(word[fout[i]].id==0)
            cout<<word[fout[i]].name<<" ";
        else
        {
            switch (word[fout[i]].id)
            {
            case 1: cout<<"and ";break;
            case 2: cout<<"or ";break;
            case 3: cout<<"not ";break;
            case 4: cout<<"> ";break;
            case 5: cout<<"< ";break;
            case 6: cout<<"= ";break;
            case 7: cout<<"!= ";break;
            case 8: cout<<">= ";break;
            case 9: cout<<"<= ";break;
            }
        }
    }
    index=0;
 }
int compare(int i)
{
    int ii;
    i=not(i);
    while(word[i].id==4||word[i].id==5||word[i].id==6
        ||word[i].id==7||word[i].id==8||word[i].id==9)
    {
        ii=i;
        i++;                
        i=not(i);
        fout[index]=ii;
        index++;
    }
    return i;
}
void main()
{
    char buffer[30];
    int i=0;
    int length;
    char *inword;
    init();
    print();
    cout<<"输入布尔表达式,以#结束 :"<<endl;
    cin>>buffer[i];
    while(buffer[i]!='#')
    {
        i++;
        cin.get(buffer[i]);
    }
    buffer[i]='\0';
    length=i+1;
    inword=new char[length];
    memcpy(inword,buffer,length);
    length=lex(inword);
    cout<<"词法分析结果如下:"<<endl;
    cout<<'('<<"属性"<<' '<<"值"<<')';
    for( i=0;i<length;i++)
    {    cout<<'('<<word[i].id<<' '<<word[i].name<<')'; cout<<' ';}
    cout<<endl;

    n=length;
    forboland(length);
    gram(k);
    cout<<"逆波兰式为:\n"<<endl;
    //逆波兰式翻译
    boland();
    cout<<endl;
}
搜索更多相关主题的帖子: int 布尔 波兰 语法分析 词法分析 
2008-06-23 11:46
zjl138
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你这样贴出来神仙也看不懂的,自已写一下不就什么都懂了.

i like linux...
2008-06-23 11:54
mcgradyzj
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看来只有我顶你啊!王鹏!!!!!
2008-06-25 21:52
晨曦的朝阳
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晕死,我扫了十秒,从上到下,呵呵!
2008-06-28 21:31
快速回复:关于 使用词法分析,语法分析将布尔表达式转换为逆波兰式 的下了个程序 ...
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