词法分析程序(Lexical Analyzer)要求:
- 从左至右扫描构成源程序的字符流
- 识别出有词法意义的单词(Lexemes)
- 返回单词记录(单词类别,单词本身)
- 滤掉空格
- 跳过注释
- 发现词法错误
程序结构:
输入:字符流(什么输入方式,什么数据结构保存)
处理:
–遍历(什么遍历方式)
–词法规则
输出:单词流(什么输出形式)
–二元组
单词类别:
1.标识符(10)
2.无符号数(11)
3.保留字(一词一码)
4.运算符(一词一码)
5.界符(一词一码)
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源代码:
#include<stdio.h>
#include<string.h>
#include<iostream>
char prog[80],token[8];
char ch;
int syn,p,m=0,n,row,sum=0;
char *word[6]={"begin","if","then","while","do","end"};
#include<string.h>
#include<iostream>
char prog[80],token[8];
char ch;
int syn,p,m=0,n,row,sum=0;
char *word[6]={"begin","if","then","while","do","end"};
void scaner()
{
for(n=0;n<8;n++) token[n]=NULL;
ch=prog[p++];
while(ch==' ')
{
ch=prog[p];
p++;
}
if((ch>='a'&&ch<='z')||(ch>='A'&&ch<='Z'))
{
m=0;
while((ch>='0'&&ch<='9')||(ch>='a'&&ch<='z')||(ch>='A'&&ch<='Z'))
{
token[m++]=ch;
ch=prog[p++];
}
token[m++]='\0';
p--;
syn=10;
for(n=0;n<6;n++)
if(strcmp(token,word[n])==0)
{
syn=n+1;
break;
}
}
else if((ch>='0'&&ch<='9'))
{
{
sum=0;
while((ch>='0'&&ch<='9'))
{
sum=sum*10+ch-'0';
ch=prog[p++];
}
}
p--;
syn=11;
if(sum>32767)
syn=-1;
}
else switch(ch)
{
case'<':m=0;token[m++]=ch;
ch=prog[p++];
if(ch=='>')
{
syn=21;
token[m++]=ch;
}
else if(ch=='=')
{
syn=22;
token[m++]=ch;
}
else
{
syn=23;
p--;
}
break;
case'>':m=0;token[m++]=ch;
ch=prog[p++];
if(ch=='=')
{
syn=24;
token[m++]=ch;
}
else
{
syn=20;
p--;
}
break;
case':':m=0;token[m++]=ch;
ch=prog[p++];
if(ch=='=')
{
syn=18;
token[m++]=ch;
}
else
{
syn=17;
p--;
}
break;
case'*':syn=13;token[0]=ch;break;
case'/':syn=14;token[0]=ch;break;
case'+':syn=15;token[0]=ch;break;
case'-':syn=16;token[0]=ch;break;
case'=':syn=25;token[0]=ch;break;
case';':syn=26;token[0]=ch;break;
case'(':syn=27;token[0]=ch;break;
case')':syn=28;token[0]=ch;break;
case'#':syn=0;token[0]=ch;break;
case'\n':syn=-2;break;
default: syn=-1;break;
}
}
int main()
{
p=0;
row=1;
printf("请输入:");
do
{
ch = getchar();
prog[p++]=ch;
}
while(ch!='#');
p=0;
do
{
scaner();
switch(syn)
{
case 11: printf("(%d,%s),syn,sum");
default: printf("(%d,%s)\n",syn,token);
}
}
while (syn!=0);
}
{
for(n=0;n<8;n++) token[n]=NULL;
ch=prog[p++];
while(ch==' ')
{
ch=prog[p];
p++;
}
if((ch>='a'&&ch<='z')||(ch>='A'&&ch<='Z'))
{
m=0;
while((ch>='0'&&ch<='9')||(ch>='a'&&ch<='z')||(ch>='A'&&ch<='Z'))
{
token[m++]=ch;
ch=prog[p++];
}
token[m++]='\0';
p--;
syn=10;
for(n=0;n<6;n++)
if(strcmp(token,word[n])==0)
{
syn=n+1;
break;
}
}
else if((ch>='0'&&ch<='9'))
{
{
sum=0;
while((ch>='0'&&ch<='9'))
{
sum=sum*10+ch-'0';
ch=prog[p++];
}
}
p--;
syn=11;
if(sum>32767)
syn=-1;
}
else switch(ch)
{
case'<':m=0;token[m++]=ch;
ch=prog[p++];
if(ch=='>')
{
syn=21;
token[m++]=ch;
}
else if(ch=='=')
{
syn=22;
token[m++]=ch;
}
else
{
syn=23;
p--;
}
break;
case'>':m=0;token[m++]=ch;
ch=prog[p++];
if(ch=='=')
{
syn=24;
token[m++]=ch;
}
else
{
syn=20;
p--;
}
break;
case':':m=0;token[m++]=ch;
ch=prog[p++];
if(ch=='=')
{
syn=18;
token[m++]=ch;
}
else
{
syn=17;
p--;
}
break;
case'*':syn=13;token[0]=ch;break;
case'/':syn=14;token[0]=ch;break;
case'+':syn=15;token[0]=ch;break;
case'-':syn=16;token[0]=ch;break;
case'=':syn=25;token[0]=ch;break;
case';':syn=26;token[0]=ch;break;
case'(':syn=27;token[0]=ch;break;
case')':syn=28;token[0]=ch;break;
case'#':syn=0;token[0]=ch;break;
case'\n':syn=-2;break;
default: syn=-1;break;
}
}
int main()
{
p=0;
row=1;
printf("请输入:");
do
{
ch = getchar();
prog[p++]=ch;
}
while(ch!='#');
p=0;
do
{
scaner();
switch(syn)
{
case 11: printf("(%d,%s),syn,sum");
default: printf("(%d,%s)\n",syn,token);
}
}
while (syn!=0);
}
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运行结果截图:
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另外附上第二次、第三次未提交的作业链接(忘记提交,但发博客日期在规定时间内)
第二次:https://www.cnblogs.com/dyun3/p/11511370.html
第三次:https://www.cnblogs.com/dyun3/p/11551622.html