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  • 实验一、词法分析器实验 199

    专业:商业软件工程   姓名:陈楷涛  学号:201506110199

    一、        实验目的

           通过编写一个词法分析器,进一步了解词法之间的关系

    二、        实验内容和要求

    对输入的字符串进行识别,并将输出该字符串和种别码

    三、        实验方法、步骤及结果测试

     1.      源程序名:词法分析.c

    可执行程序名:词法分析.exe

    2.      原理分析及流程图

     

    #include <stdio.h>

    #include <string.h>

    char prog[80],token[8],ch;

    int syn,p,m,n,sum;

    char *rwtab[6]={"begin","if","then","while","do","end"};

    void scaner(void);

     main()          主函数:循坏输入字符串和输出最后的字符串和识别码

     {

         p=0;

        printf("\n please input a string(end with '#'):\n");

           do{

                  scanf("%c",&ch);

                  prog[p++]=ch;

          }while(ch!='#');

           p=0;

         do{

                  scaner();        调用scaner函数,循环输出字符串和识别码

                  switch(syn)

                   {

                       case 11:

                           printf("( %-10d%5d )\n",sum,syn);

                        break;

                        case -1:

                            printf("you have input a wrong string\n");

                            return 0;

                        break;

                        default:

                        printf("( %-10s%5d )\n",token,syn);

                        break;

                    }

             }while(syn!=0);

       }

      void scaner(void)     子函数scaner:对输入的字符串进行分析

      {

           sum=0;

          for(m=0;m<8;m++)

              token[m++]= NULL;

              ch=prog[p++];

              m=0;

          while((ch==' ')||(ch=='\n'))

              ch=prog[p++];

          if(((ch<='z')&&(ch>='a'))||((ch<='Z')&&(ch>='A')))

           {

                while(((ch<='z')&&(ch>='a'))||((ch<='Z')&&(ch>='A'))||((ch>='0')&&(ch<='9')))

                {

                    token[m++]=ch;

                     ch=prog[p++];

               }

                p--;

                syn=10;

                for(n=0;n<6;n++)

              if(strcmp(token,rwtab[n])==0)

                 {

                     syn=n+1;

                  break;

                 }

           }

           else if((ch>='0')&&(ch<='9'))

           {

                while((ch>='0')&&(ch<='9'))

              {

                  sum=sum*10+ch-'0';

                    ch=prog[p++];

              }

                 syn=11;

          }

          else

          {

              switch(ch)

              {

              case '<':

                  token[m++]=ch;

                  ch=prog[p++];

                  if(ch=='=')

                  {

                      syn=22;

                      token[m++]=ch;

                  }

                  else

                  {

                      syn=20;

                      p--;

                  }

              break;

              case '>':

                  token[m++]=ch;

                  ch=prog[p++];

                  if(ch=='=')

                  {

                      syn=24;

                      token[m++]=ch;

                  }

                  else

                  {

                      syn=23;

                      p--;

                  }

             break;

             case '+':

                 token[m++]=ch;

                 ch=prog[p++];

                 if(ch=='+')

                 {

                     syn=17;

                     token[m++]=ch;

                 }

                 else

                {

                     syn=13;

                     p--;

                 }

             break;

             case '-':

                 token[m++]=ch;

                 ch=prog[p++];

                 if(ch=='-')

                 {

                    syn=29;

                     token[m++]=ch;

                }

                 else

                 {

                     syn=14;

                     p--;

                 }

             break;

             case '!':

                 ch=prog[p++];

                 if(ch=='=')

                 {

                     syn=21;

                     token[m++]=ch;

                 }

                 else

                 {

                     syn=31;

                     p--;

                 }

             break;

             case '=':

                 token[m++]=ch;

                 ch=prog[p++];

                 if(ch=='=')

                 {

                     syn=25;

                     token[m++]=ch;

                 }

                 else

                 {

                     syn=18;

                     p--;

                 }

             break;

             case '*':

                 syn=15;

                 token[m++]=ch;

             break;

             case '/':

                 syn=16;

                 token[m++]=ch;

             break;

             case '(':

                 syn=27;

                 token[m++]=ch;

             break;

             case ')':

                 syn=28;

                 token[m++]=ch;

             break;

             case '{':

                 syn=5;

                 token[m++]=ch;

             break;

             case '}':

                 syn=6;

                 token[m++]=ch;

             break;

             case ';':

                 syn=26;

                 token[m++]=ch;

             break;

             case '\"':

                 syn=30;

                 token[m++]=ch;

             break;

             case '#':

                 syn=0;

                token[m++]=ch;

             break;

             case ':':

                syn=17;

                 token[m++]=ch;

             break;

             default:

                 syn=-1;

             break;

             }

         }

             token[m++]='\0';

     }

    4.      运行结果及分析

    四、        实验总结

    心得体会:这是我第一次用C语言编写这么一个词法分析的工具,在编写这个词法分析器的过程中,我感觉我对词法分析也有了更深入的了解,但也不可避免的出现了一些问题,例如,如何原字符串进行拆分,然后一个一个进行分析 一度成为这个程序的障碍,最后,我采用了数组的方式来解决了这个问题。总而言之,这个程序为我打开了一项从未涉及过的崭新的大门,我将在这座大门后面开始我新的学习旅程。

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  • 原文地址:https://www.cnblogs.com/cktcom/p/5955030.html
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