Question

求一份C语言编写的极其简化的C词法分析器，高手应该不超过20分钟就能写出来

几个月都没编程了，过几天就要交了，很是头疼
要求很简单：1.识别C常见关键字（10个就够了，随便定义10个常见的关键字），常量（只识别0-100的整数）和运算符（+ - / *）
2. 运行时手动输入一段C语言源程序，结果将识别出来的关键字，常量（整数0-100），和运算符分别显示出来，不需要编号和统计数目。运行格式如下显示：
关键字：if then and array bool等
常量：5 36 15 48等
运算符：+ - * 等

Answer 1

#include <iostream>

#include<string>

using namespace std;

#define MAX 22

char ch =' ';

string key[15]={"begin","end","if","then","else","while","write","read",

"do", "call","const","char","碰带until","procedure","repeat"};

int Iskey(string c){ //关键字判断

int i;

for(i=0;i<MAX;i++) {

if(key[i].compare(c)==0) return 1;

}

return 0;

}

int IsLetter(char c) { //判断是否为字母

if(((c<='z')&&(c>='a'))||((c<='Z')&&(c>='A'))) return 1;

else return 0;

}

int IsDigit(char c){ //判断是否为数字

if(c>='0'&&c<='9') return 1;

else return 0;

}

void analyse(FILE *fpin){

string arr="";

while((ch=fgetc(fpin))!=EOF) {

arr="";

if(ch==' '||ch=='\t'||ch=='\n'搏吵灶){}

else if(IsLetter(ch)){

while(IsLetter(ch)||IsDigit(ch)) {

if((ch<='Z')&&(ch>='A')) ch=ch+32;

arr=arr+ch;

ch=fgetc(fpin);

}

fseek(fpin,-1L,SEEK_CUR);

if (Iskey(arr)){cout<<arr<<"\t$关键字"<<endl;}

else cout<<arr<<"\t$普通标识符"<<endl;

}

else if(IsDigit(ch)){

while(IsDigit(ch)||ch=='.'&&IsDigit(fgetc(fpin))){

arr=arr+ch;

ch=fgetc(fpin);

}

fseek(fpin,-1L,SEEK_CUR);

cout<<arr<<"\t$无符号实数"<<endl;

}

else switch(ch){

case'+':

case'-' :

case'*' :

case'=' :

case'/' :cout<<ch<<"\t$运算符"<<endl;break;

case'(' :

case')' :

case'[' :

case']' :

case';' :

case'.' :

case',' :

case'{' :

case'}' :cout<<ch<<"\t$界符"<<endl;break;

case':' :{ch=fgetc(fpin);

if(ch=='基扮=') cout<<":="<<"\t$运算符"<<endl;

else {cout<<"="<<"\t$运算符"<<endl;;

fseek(fpin,-1L,SEEK_CUR);}

}break;

case'>' :{ch=fgetc(fpin);

if(ch=='=') cout<<">="<<"\t$运算符"<<endl;

if(ch=='>')cout<<">>"<<"\t$输入控制符"<<endl;

else {cout<<">"<<"\t$运算符"<<endl;

fseek(fpin,-1L,SEEK_CUR);}

}break;

case'<' :{ch=fgetc(fpin);

if(ch=='=')cout<<"<="<<"\t$运算符"<<endl;

else if(ch=='<')cout<<"<<"<<"\t$输出控制符"<<endl;

else if(ch=='>') cout<<"<>"<<"\t$运算符"<<endl;

else{cout<<"<"<<"\t$运算符"<<endl;

fseek(fpin,-1L,SEEK_CUR);}

}break;

default : cout<<ch<<"\t$无法识别字符"<<endl;

}

}

}

void main(){

char in_fn[30];

FILE * fpin;

cout<<"请输入源文件名（包括路径和后缀名）:";

for(;;){

cin>>in_fn;

if((fpin=fopen(in_fn,"r"))!=NULL) break;

else cout<<"文件路径错误！请输入源文件名（包括路径和后缀名）:";

}

cout<<"\n********************分析如下*********************"<<endl;

analyse(fpin);

fclose(fpin);

}

Answer 2

呵呵 我们刚刚做了这个交了 差别还是有点大的。不知道你要吗，要的话追问我。

追问

可不可以把源代码发出来？

追答

#include
#include
#include
#define MAX 256
char baoliuzi[10][10] =
{
"main","if","else","switch","int","do","while","char","write","read"//保留字
};
/*打开对应的代码文件，2元素文件，标识符文件，常熟文件*/
FILE *fp,*wfp;
int id_count = 0;
int num_count = 0;
char word[MAX],ch;
char a[20],b[20];//原文件路径和保存路径
int n_word = 0;
/*对应保留字的编码*/
int c;
/*finished read the file*/
int over = 0;
/*扫描指针回退一个字符*/
void retract()
{
fseek(fp,-1,SEEK_CUR);
}

int init()
{
fp = fopen(a,"a");
if (fp==NULL)
{
printf("打开失败。\n");
return 0;
}
fprintf(fp,"%c",'\n');//将文件末尾添加结束符
fclose(fp);
}
/*将当前读入的字符送入word数组中*/
void concatenation()
{
word[n_word] = ch;
n_word++;
word[n_word] = '\0';
}
/*判断是否是字母*/
int letter()
{
if ((ch >='a'&&ch='A'&&ch='a'&&ch='A'&&ch='0'&&ch ='0'&&ch <='9')
return 1;
else
return 0;
}
/*判断是否是保留字。是则返回保留字对应的编码不是则返回标识符编码6*/
int reserve()
{
if (strcmp(word,"main") == 0)
return 1;
if (strcmp(word,"if") == 0)
return 2;
if (strcmp(word,"else") == 0)
return 3;
if (strcmp(word,"switch") == 0)
return 4;
if (strcmp(word,"int") == 0)
return 5;
if (strcmp(word,"do") == 0)
return 6;
if (strcmp(word,"while") == 0)
return 7;
if (strcmp(word,"char") == 0)
return 8;
if (strcmp(word,"write") == 0)
return 9;
if (strcmp(word,"read") == 0)
return 10;
return 0;
}
void error()
{
fprintf(wfp,"%s\t%c\n","Error",ch);
}
主函数在下面

Answer 3

我不是高手