《編譯原理》實驗2

1、編譯原理（實驗部分）實驗 2_PL0 詞法分析一、實驗目的加深和鞏固對于詞法分析的了解和掌握；初步認識PL/0 語言的基礎和簡單的程序編寫；通過本實驗能夠初步的了解和掌握程序詞法分析的整個過程；提高自己上機和編程過程中處理具體問題的能力。二、實驗設備1、PC 兼容機一臺；操作系統(tǒng)為WindowsWindowsXP。2、 Visual C+ 6.0或以上版本，Windows 2000 或以上版本，匯編工具（在 Software子目錄下）。三、實驗原理PL/O 語言的編譯程序，是用高級語言PASCAL語言書寫的。整個編譯過程是由一些嵌套及并列的過程或函數(shù)完成。詞法分析程序是獨立的過程GETSYM完

2、成，供語法分析讀單詞時使用。四、實驗步驟閱讀所給出的詞法分析程序（pl0_lexical.c），搞懂程序中每一個變量的含義，以及每一個過程的作用，并在該過程中進行中文注釋；閱讀完程序后，畫出各過程的流程圖；給出的程序包含兩處輸入錯誤，利用所給的pl/0源程序(test.pl0)對程序進行調(diào)試，使其能正確對所給文件進行分析并能夠解釋運行；在閱讀懂所給出的詞法分析程序后，將你對詞法分析的理解寫在實驗報告上。實驗代碼#include<stdio.h>main() printf ("my name is 08061118 yuchaofeng ");主程序：#inclu

3、de <stdio.h>#include <ctype.h>#include <alloc.h>#include <stdlib.h>#include <string.h>#define NULL 0FILE *fp;char cbuffer;char *key8="DO","BEGIN","ELSE","END","IF","THEN","VAR","WHILE"char *

4、border6=",","",":=",".","(",")"char *arithmetic4="+","-","*","/"char *relation6="<","<=","=",">",">=","<>"char *consts20;c

5、har *label20;int constnum=0,labelnum=0;int search(char searchchar,int wordtype)int i=0;switch (wordtype) case 1:for (i=0;i<=7;i+)if (strcmp(keyi,searchchar)=0)return(i+1);case 2:for (i=0;i<=5;i+)if (strcmp(borderi,searchchar)=0)return(i+1);return(0);case 3:for (i=0;i<=3;i+) if (strcmp(arith

6、metici,searchchar)=0)return(i+1);return(0);case 4:for (i=0;i<=5;i+) if (strcmp(relationi,searchchar)=0)return(i+1);return(0);case 5:for (i=0;i<=constnum;i+)if (strcmp(constsi,searchchar)=0)return(i+1);constsi-1=(char *)malloc(sizeof(searchchar);strcpy(constsi-1,searchchar);constnum+;return(i);

7、case 6:for (i=0;i<=labelnum;i+) if (strcmp(labeli,searchchar)=0)return(i+1);labeli-1=(char *)malloc(sizeof(searchchar);strcpy(labeli-1,searchchar);labelnum+;return(i);char alphaprocess(char buffer)int atype;int i=-1;char alphatp20;while (isalpha(buffer)|(isdigit(buffer) alphatp+i=buffer; buffer=f

8、getc(fp);alphatpi+1='0'if (atype=search(alphatp,1)printf("%s (1,%d)n",alphatp,atype-1);elseatype=search(alphatp,6);printf("%s (6,%d)n",alphatp,atype-1);return(buffer);char digitprocess(char buffer)int i=-1;char digittp20;int dtype;while (isdigit(buffer) digittp+i=buffer;

9、buffer=fgetc(fp);digittpi+1='0'dtype=search(digittp,5);printf("%s (5,%d)n",digittp,dtype-1);return(buffer);char otherprocess(char buffer)int i=-1;char othertp20;int otype,otypetp;othertp0=buffer;othertp1='0'if (otype=search(othertp,3)printf("%s (3,%d)n",othertp,ot

10、ype-1);buffer=fgetc(fp);goto out;if (otype=search(othertp,4)buffer=fgetc(fp);othertp1=buffer;othertp2='0'if (otypetp=search(othertp,4)printf("%s (4,%d)n",othertp,otypetp-1);goto out;elseothertp1='0'printf("%s (4,%d)n",othertp,otype-1);goto out;if (buffer=':

11、9;) buffer=fgetc(fp); if (buffer='=')printf(":= (2,2)n");buffer=fgetc(fp);goto out;elseif (otype=search(othertp,2) printf("%s (2,%d)n",othertp,otype-1); buffer=fgetc(fp);goto out;if (buffer!='n')&&(buffer!=' ')printf("%c error,not a wordn",buffer);buffer=fgetc(fp);out:return(buffer);void main()int i;for (i=0;i<=20;i+)labeli=NULL;constsi=NULL;if (fp=fopen("skh.c","r")=NULL)printf("error");elsecbuffer = fgetc(fp);wh