Question

词法分析问题

求一段用c++写的词法分析的程序，功能只要一个就是提取C程序的变量，要源程序！真是写不出来了，高手进，急急急急急急急急急急急急

Answer 1

#include<iostream.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int i,j,k,flag,number,status;
/*status which is use to judge the string is keywords or not!*/
char ch;
char words[10] = {" "};
char program[500];
int Scan(char program[])
{
char *keywords[13] = {"void","main","if","then","break","int",
"char","float","include","for","while","printf",
"scanf"};
number = 0;
status = 0;
j = 0;
ch = program[i++];
/* To handle the lettle space ands tab*/
/*handle letters*/
if ((ch >= 'a') && (ch <= 'z' ))
{
while ((ch >= 'a') && (ch <= 'z' ))
{
words[j++]=ch;
ch=program[i++];
}
i--;
words[j++] = '\0';
for (k = 0; k < 13; k++)
if (strcmp (words,keywords[k]) == 0)
switch(k)
{
case 0:{
flag = 1;
status = 1;
break;
}
case 1:{
flag = 2;
status = 1;
break;
}
case 2:{
flag = 3;
status = 1;
break;
}
case 3:{
flag = 4;
status = 1;
break;
}
case 4:{
flag = 5;
status = 1;
break;
}
case 5:{
flag = 6;
status = 1;
break;
}
case 6:{
flag = 7;
status = 1;
break;
}
case 7:{
flag = 8;
status = 1;
break;
}
case 8:{
flag = 9;
status = 1;
break;
}
case 9:{
flag = 10;
status = 1;
break;
}
case 10:{
flag = 11;
status = 1;
break;
}
case 11:{
flag = 12;
status = 1;
break;
}
case 12:{
flag = 13;
status = 1;
break;
}
}
if (status == 0)
{
flag = 100;
}
}
/*handle digits*/
else if ((ch >= '0') && (ch <= '9'))
{
number = 0;
while ((ch >= '0' ) && (ch <= '9' ))
{
number = number*10+(ch-'0');
ch = program[i++];
}
flag = 200;
i--;
}
/*opereation and edge handle*/
else switch (ch)
{
case '=':{
if (ch == '=')
words[j++] = ch;
words[j] = '\0';
ch = program[i++];
if (ch == '=')
{
words[j++] = ch;
words[j] = '\0';
flag = 401;
}
else
{
i--;
flag = 402;
}
break;
}
case'>':{
if (ch == '>')
words[j++] = ch;
words[j] = '\0';
ch = program[i++];
if (ch == '=')
{
words[j++] = ch;
words[j] = '\0';
flag = 403;
}
else
{
i--;
flag = 404;
}
break;
}
case'<':{
if (ch == '<')
words[j++] = ch;
words[j] = '\0';
ch = program[i++];
if (ch == '=')
{
words[j++] = ch;
words[j] = '\0';
flag = 405;
}
else
{
i--;
flag = 406;
}
break;
}
case'!':{
if (ch == '!')
words[j++] = ch;
words[j] = '\0';
ch = program[i++];
if (ch == '=')
{
words[j++] = ch;
words[j] = '\0';
flag = 407;
}
else
{
i--;
flag = 408;
}
break;
}
case'+':{
if (ch == '+')
words[j++] = ch;
words[j] = '\0';
ch = program[i++];
if (ch == '=')
{
words[j++] = ch;
words[j] = '\0';
flag = 409;
}
else if (ch == '+')
{
words[j++] = ch;
words[j] = '\0';
flag = 410;
}
else
{
i--;
flag = 411;
}
break;
}
case'-':{
if (ch == '-')
words[j++] = ch;
words[j] = '\0';
ch = program[i++];
if (ch == '=')
{
words[j++] = ch;
words[j] = '\0';
flag = 412;
}
else if( ch == '-')
{
words[j++] = ch;
words[j] = '\0';
flag = 413;
}
else
{
i--;
flag = 414;
}
break;
}
case'*':{
if (ch == '*')
words[j++] = ch;
words[j] = '\0';
ch = program[i++];
if (ch == '=')
{
words[j++] = ch;
words[j] = '\0';
flag = 415;
}
else
{
i--;
flag = 416;
}
break;
}
case'/':{
if (ch == '/')
words[j++] = ch;
words[j] = '\0';
ch = program[i++];
if (ch == '=')
{
words[j++] = ch;
words[j] = '\0';
flag = 417;
}
else
{
i--;
flag = 418;
}
break;
}
case';':{
words[j] = ch;
words[j+1] = '\0';
flag = 501;
break;
}
case'(':{
words[j] = ch;
words[j+1] = '\0';
flag = 502;
break;
}
case')':{
words[j] = ch;
words[j+1] = '\0';
flag = 503;
break;
}
case'[':{
words[j] = ch;
words[j+1] = '\0';
flag = 504;
break;
}
case']':{
words[j] = ch;
words[j+1] = '\0';
flag = 505;
break;
}
case'{':{
words[j] = ch;
words[j+1] = '\0';
flag = 506;
break;
}
case'}':{
words[j] = ch;
words[j+1] = '\0';
flag = 507;
break;
}
case':':{
words[j] = ch;
words[j+1] = '\0';
flag = 508;
break;
}
case'"':{
words[j] = ch;
words[j+1] = '\0';
flag = 509;
break;
}
case'%':{
if (ch == '%')
words[j++] = ch;
words[j] = '\0';
ch = program[i++];
if (ch == '=')
{
words[j++] = ch;
words[j] = '\0';
flag = 510;
}
else
{
i--;
flag = 511;
}
break;
}
case',':{
words[j] = ch;
words[j+1] = '\0';
flag = 512;
break;
}
case'#':{
words[j] = ch;
words[j+1] = '\0';
flag = 513;
break;
}
case'@':{
words[j] = '#';
flag = 0;
break;
}
default:{
flag = -1;
break;
}
}
return flag;
}
main()
{
i=0;
printf("please input a program end with @");
do
{
ch = getchar();
program[i++] = ch;
}while(ch != '@');
i = 0;
do{
flag = Scan(program);
if (flag == 200)
{
printf("(%2d,%4d)",flag,number);
}
else if (flag == -1)
{
printf("(%d,error)",flag);
}
else
{
printf("(%2d,%4s)",flag,words);
}
}while (flag != 0);
system("pause");
}

Answer 2

case语句的格式是“的情??况下（空间）的表情：”寻求执行，加上括号中的语句的case语句后面的顺。
上面的程序，少了两个括号
情况下，'>'：{
in.get（CH）;
（CH =='='）出来“” > =“<<”“<<”运算符“<< endl;
（CH =='>'）了”;“>>”<<“”<< endl <其他{出<”<“<<”操作符“<< endl;
}打破;
情况下，”<“：{
in.get（CH）
（CH =='='）出来<<“=”<<“”<<“操作符”<< endl;
其他（CH =='<'）出<“<<”<<“”<< endl <<分隔符;
其他{出<<“”<<“”<<“操作符”<< endl;
};