編譯原理詞法分析實驗怎麼做
❶ 有人知道編譯原理實驗之詞法分析器用C++怎麼做嗎
#include "globals.h"
#include "util.h"
#include "scan.h"
#include "parse.h"
static TokenType token; /* holds current token */
/* function prototypes for recursive calls */
static TreeNode * stmt_sequence(void);
static TreeNode * statement(void);
static TreeNode * if_stmt(void);
static TreeNode * repeat_stmt(void);
static TreeNode * assign_stmt(void);
static TreeNode * read_stmt(void);
static TreeNode * write_stmt(void);
static TreeNode * exp(void);
static TreeNode * simple_exp(void);
static TreeNode * term(void);
static TreeNode * factor(void);
static void syntaxError(char * message)
{ fprintf(listing,"\n>>> ");
fprintf(listing,"Syntax error at line %d: %s",lineno,message);
Error = TRUE;
}
static void match(TokenType expected)
{ if (token == expected) token = getToken();
else {
syntaxError("unexpected token -> ");
printToken(token,tokenString);
fprintf(listing," ");
}
}
TreeNode * stmt_sequence(void)
{ TreeNode * t = statement();
TreeNode * p = t;
while ((token!=ENDFILE) && (token!=END) &&
(token!=ELSE) && (token!=UNTIL))
{ TreeNode * q;
match(SEMI);
q = statement();
if (q!=NULL) {
if (t==NULL) t = p = q;
else /* now p cannot be NULL either */
{ p->sibling = q;
p = q;
}
}
}
return t;
}
TreeNode * statement(void)
{ TreeNode * t = NULL;
switch (token) {
case IF : t = if_stmt(); break;
case REPEAT : t = repeat_stmt(); break;
case ID : t = assign_stmt(); break;
case READ : t = read_stmt(); break;
case WRITE : t = write_stmt(); break;
default : syntaxError("unexpected token -> ");
printToken(token,tokenString);
token = getToken();
break;
} /* end case */
return t;
}
TreeNode * if_stmt(void)
{ TreeNode * t = newStmtNode(IfK);
match(IF);
if (t!=NULL) t->child[0] = exp();
match(THEN);
if (t!=NULL) t->child[1] = stmt_sequence();
if (token==ELSE) {
match(ELSE);
if (t!=NULL) t->child[2] = stmt_sequence();
}
match(END);
return t;
}
TreeNode * repeat_stmt(void)
{ TreeNode * t = newStmtNode(RepeatK);
match(REPEAT);
if (t!=NULL) t->child[0] = stmt_sequence();
match(UNTIL);
if (t!=NULL) t->child[1] = exp();
return t;
}
TreeNode * assign_stmt(void)
{ TreeNode * t = newStmtNode(AssignK);
if ((t!=NULL) && (token==ID))
t->attr.name = String(tokenString);
match(ID);
match(ASSIGN);
if (t!=NULL) t->child[0] = exp();
return t;
}
TreeNode * read_stmt(void)
{ TreeNode * t = newStmtNode(ReadK);
match(READ);
if ((t!=NULL) && (token==ID))
t->attr.name = String(tokenString);
match(ID);
return t;
}
TreeNode * write_stmt(void)
{ TreeNode * t = newStmtNode(WriteK);
match(WRITE);
if (t!=NULL) t->child[0] = exp();
return t;
}
TreeNode * exp(void)
{ TreeNode * t = simple_exp();
if ((token==LT)||(token==EQ)) {
TreeNode * p = newExpNode(OpK);
if (p!=NULL) {
p->child[0] = t;
p->attr.op = token;
t = p;
}
match(token);
if (t!=NULL)
t->child[1] = simple_exp();
}
return t;
}
TreeNode * simple_exp(void)
{ TreeNode * t = term();
while ((token==PLUS)||(token==MINUS))
{ TreeNode * p = newExpNode(OpK);
if (p!=NULL) {
p->child[0] = t;
p->attr.op = token;
t = p;
match(token);
t->child[1] = term();
}
}
return t;
}
TreeNode * term(void)
{ TreeNode * t = factor();
while ((token==TIMES)||(token==OVER))
{ TreeNode * p = newExpNode(OpK);
if (p!=NULL) {
p->child[0] = t;
p->attr.op = token;
t = p;
match(token);
p->child[1] = factor();
}
}
return t;
}
TreeNode * factor(void)
{ TreeNode * t = NULL;
switch (token) {
case NUM :
t = newExpNode(ConstK);
if ((t!=NULL) && (token==NUM))
t->attr.val = atoi(tokenString);
match(NUM);
break;
case ID :
t = newExpNode(IdK);
if ((t!=NULL) && (token==ID))
t->attr.name = String(tokenString);
match(ID);
break;
case LPAREN :
match(LPAREN);
t = exp();
match(RPAREN);
break;
default:
syntaxError("unexpected token -> ");
printToken(token,tokenString);
token = getToken();
break;
}
return t;
}
/****************************************/
/* the primary function of the parser */
/****************************************/
/* Function parse returns the newly
* constructed syntax tree
*/
TreeNode * parse(void)
{ TreeNode * t;
token = getToken();
t = stmt_sequence();
if (token!=ENDFILE)
syntaxError("Code ends before file\n");
return t;
}
上面是一個語法分析器的主代碼部分它可以識別類似下面的代碼，但是由於篇幅有限，上面的代碼不是完整代碼，完整代碼太長，還有好幾個文件。
read x; { input an integer }
if 0 < x then { don't compute if x <= 0 }
fact := 1;
repeat
fact := fact * x;
x := x - 1
until x = 0;
write fact { output factorial of x }
end
❷ 編譯原理詞法分析
編譯的詞法分析，一般是先畫一個狀態轉換圖，一般是有多少分支，就有多少if語句，分支裡面再分（可能有循環語句）。注意記住詞的類別和詞的字元串，請以以下代碼為例，理會一下詞法分析的大致過程。
while(s[i]!='#')
{
while(s[i]==' '||s[i]=='\t'||s[i]=='\n')
{
if(s[i]=='\n')
line++;
i++;
}
if(s[i]=='#')
break;
j=i;
if(s[i]>='a'&&s[i]<='z'||s[i]>='A'&&s[i]<='Z')
{
i++;
while(s[i]>='a'&&s[i]<='z'||s[i]>='A'&&s[i]<='Z'||s[i]>='0'&&s[i]<='9')
i++;
if((i-j)==2&&s[j]=='i'&&s[j+1]=='f')
{
strcpy(dancishuzu[dancigeshu].name,"if");
dancishuzu[dancigeshu].bianhao=4;
dancigeshu++;
}
else if((i-j)==3&&s[j]=='i'&&s[j+1]=='n'&&s[j+2]=='t')
{
strcpy(dancishuzu[dancigeshu].name,"int");
dancishuzu[dancigeshu].bianhao=2;
dancigeshu++;
}
else if((i-j)==3&&s[j]=='f'&&s[j+1]=='o'&&s[j+2]=='r')
{
strcpy(dancishuzu[dancigeshu].name,"for");
dancishuzu[dancigeshu].bianhao=6;
dancigeshu++;
}
else if((i-j)==4&&s[j]=='m'&&s[j+1]=='a'&&s[j+2]=='i'&&s[j+3]=='n')
{
strcpy(dancishuzu[dancigeshu].name,"main");
dancishuzu[dancigeshu].bianhao=1;
dancigeshu++;
}
else if ((i-j)==4&&s[j]=='c'&&s[j+1]=='h'&&s[j+2]=='a'&&s[j+3]=='r')
{
strcpy(dancishuzu[dancigeshu].name,"char");
dancishuzu[dancigeshu].bianhao=3;
dancigeshu++;
}
else if ((i-j)==4&&s[j]=='e'&&s[j+1]=='l'&&s[j+2]=='s'&&s[j+3]=='e')
{
strcpy(dancishuzu[dancigeshu].name,"else");
dancishuzu[dancigeshu].bianhao=5;
dancigeshu++;
}
else if ((i-j)==5&&s[j]=='w'&&s[j+1]=='h'&&s[j+2]=='i'&&s[j+3]=='l'&&s[j+4]=='e')
{
strcpy(dancishuzu[dancigeshu].name,"while");
dancishuzu[dancigeshu].bianhao=7;
dancigeshu++;
}
else{
dancishuzu[dancigeshu].bianhao=10;
count=0;
while(j<i)
{
dancishuzu[dancigeshu].name[count++]=s[j];
j++;
}
dancishuzu[dancigeshu].name[count]='\0';
dancigeshu++;
}
}
else if(s[i]>='0'&&s[i]<='9')
{
while(s[i]>='0'&&s[i]<='9')
i++;
dancishuzu[dancigeshu].bianhao=11;
count=0;
while(j<i)
{
dancishuzu[dancigeshu].name[count++]=s[j];
j++;
}
dancishuzu[dancigeshu].name[count]='\0';
dancigeshu++;
}
else if(s[i]=='=')
{
if(s[i+1]=='=')
{
dancishuzu[dancigeshu].bianhao=30;
strcpy(dancishuzu[dancigeshu].name,"==");
dancigeshu++;
i+=2;
}
else
{
dancishuzu[dancigeshu].bianhao=12;
strcpy(dancishuzu[dancigeshu].name,"=");
dancigeshu++;
i++;
}
}
else if(s[i]=='+')
{
dancishuzu[dancigeshu].bianhao=13;
strcpy(dancishuzu[dancigeshu].name,"+");
dancigeshu++;
i++;
}
else if(s[i]=='-')
{
dancishuzu[dancigeshu].bianhao=14;
strcpy(dancishuzu[dancigeshu].name,"-");
dancigeshu++;
i++;
}
else if(s[i]=='*')
{
dancishuzu[dancigeshu].bianhao=15;
strcpy(dancishuzu[dancigeshu].name,"*");
dancigeshu++;
i++;
}
else if(s[i]=='/')
{
dancishuzu[dancigeshu].bianhao=16;
strcpy(dancishuzu[dancigeshu].name,"/");
dancigeshu++;
i++;
}
else if(s[i]=='(')
{
i++;
dancishuzu[dancigeshu].bianhao=17;
strcpy(dancishuzu[dancigeshu].name,"(");
dancigeshu++;
}
else if(s[i]==')')
{
i++;
dancishuzu[dancigeshu].bianhao=18;
strcpy(dancishuzu[dancigeshu].name,")");
dancigeshu++;
}
else if(s[i]=='[')
{
i++;
dancishuzu[dancigeshu].bianhao=19;
strcpy(dancishuzu[dancigeshu].name,"[");
dancigeshu++;
}
else if(s[i]==']')
{
i++;
dancishuzu[dancigeshu].bianhao=20;
strcpy(dancishuzu[dancigeshu].name,"]");
dancigeshu++;
}
else if(s[i]=='{')
{
i++;
dancishuzu[dancigeshu].bianhao=21;
strcpy(dancishuzu[dancigeshu].name,"{");
dancigeshu++;
}
else if(s[i]=='}')
{
i++;
dancishuzu[dancigeshu].bianhao=22;
strcpy(dancishuzu[dancigeshu].name,"}");
dancigeshu++;
}
else if(s[i]==',')
{
i++;
dancishuzu[dancigeshu].bianhao=23;
strcpy(dancishuzu[dancigeshu].name,",");
dancigeshu++;
}
else if(s[i]==':')
{
i++;
dancishuzu[dancigeshu].bianhao=24;
strcpy(dancishuzu[dancigeshu].name,":");
dancigeshu++;
}
else if(s[i]==';')
{
i++;
dancishuzu[dancigeshu].bianhao=25;
strcpy(dancishuzu[dancigeshu].name,";");
dancigeshu++;
}
else if(s[i]=='>')
{
if(s[i+1]=='=')
{
dancishuzu[dancigeshu].bianhao=28;
strcpy(dancishuzu[dancigeshu].name,">=");
dancigeshu++;
i+=2;
}
else
{
i++;
dancishuzu[dancigeshu].bianhao=26;
strcpy(dancishuzu[dancigeshu].name,">");
dancigeshu++;
}
}
else if(s[i]=='<')
{
if(s[i+1]=='=')
{
dancishuzu[dancigeshu].bianhao=29;
strcpy(dancishuzu[dancigeshu].name,"<=");
dancigeshu++;
i+=2;
}
else
{
i++;
dancishuzu[dancigeshu].bianhao=27;
strcpy(dancishuzu[dancigeshu].name,"<");
dancigeshu++;
}
}
else if(s[i]=='!'&&s[i+1]=='=')
{
dancishuzu[dancigeshu].bianhao=31;
strcpy(dancishuzu[dancigeshu].name,"!=");
dancigeshu++;
i+=2;
}
else
{
printf("\nline:%derror!",line);
i++;
return;
}
}
❸ 編譯原理 詞法分析程序的設計與實現實驗題
說他像蒼蠅，是罵蒼蠅呢還是罵他呢？
❹ 如何通俗易懂地解釋編譯原理中語法分析的過程
語法分析（Syntax analysis或Parsing）和語法分析程序（Parser）
語法分析是編譯過程的一個邏輯階段。語法分析的任務是在詞法分析的基礎上將單詞序列組合成各類語法短語，如「程序」，「語句」，「表達式」等等.語法分析程序判斷源程序在結構上是否正確.源程序的結構由上下文無關文法描述.
❺ 編譯原理實驗求助
1)定義
所有token或者叫單詞的有限自動機。
2)將有限自動機用代碼實現。
3)寫分析程序，利用你定義的有限自動機來識別所有的「單詞」。並將識別出來的單詞的相關信息，如名稱，位置，類別等記錄在相關的數據結構中。
❻ 編譯原理課程設計－詞法分析器設計（C語言）
#include"stdio.h"/*定義I/O庫所用的某些宏和變數*/
#include"string.h"/*定義字元串庫函數*/
#include"conio.h"/*提供有關屏幕窗口操作函數*/
#include"ctype.h"/*分類函數*/
charprog[80]={' '},
token[8];/*存放構成單詞符號的字元串*/
charch;
intsyn,/*存放單詞字元的種別碼*/
n,
sum,/*存放整數型單詞*/
m,p;/*p是緩沖區prog的指針，m是token的指針*/
char*rwtab[6]={"begin","if","then","while","do","end"};
voidscaner(){
m=0;
sum=0;
for(n=0;n<8;n++)
token[n]=' ';
ch=prog[p++];
while(ch=='')
ch=prog[p++];
if(isalpha(ch))/*ch為字母字元*/{
while(isalpha(ch)||isdigit(ch))/*ch為字母字元或者數字字元*/{
token[m++]=ch;
ch=prog[p++];}
token[m++]=' ';
ch=prog[p--];
syn=10;
for(n=0;n<6;n++)
if(strcmp(token,rwtab[n])==0)/*字元串的比較*/{
syn=n+1;
break;}}
else
if(isdigit(ch))/*ch是數字字元*/{
while(isdigit(ch))/*ch是數字字元*/{
sum=sum*10+ch-'0';
ch=prog[p++];}
ch=prog[p--];
syn=11;}
else
switch(ch){
case'<':m=0;token[m++]=ch;ch=prog[p++];
if(ch=='>'){
syn=21;
token[m++]=ch;}
elseif(ch=='='){
syn=22;
token[m++]=ch;}
else{
syn=20;
ch=prog[p--];}
break;
case'>':m=0;token[m++]=ch;ch=prog[p++];
if(ch=='='){
syn=24;
token[m++]=ch;}
else{
syn=23;
ch=prog[p--];}
break;
case':':m=0;token[m++]=ch;ch=prog[p++];
if(ch=='='){
syn=18;
token[m++]=ch;}
else{
syn=17;
ch=prog[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;
default:syn=-1;}}
main()
{
printf(" Thesignificanceofthefigures: "
"1.figures1to6saidKeyword "
"2. "
"3.figures13to28saidOperators ");
p=0;
printf(" pleaseinputstring: ");
do{
ch=getchar();
prog[p++]=ch;
}while(ch!='#');
p=0;
do{
scaner();
switch(syn){
case11:printf("(%d,%d) ",syn,sum);break;
case-1:printf(" ERROR; ");break;
default:printf("(%d,%s) ",syn,token);
}
}while(syn!=0);
getch();
}
程序測試結果
對源程序beginx:=9:ifx>9thenx:=2*x+1/3;end#的源文件，經過詞法分析後輸出如下圖5-1所示：
具體的你在修改修改吧
❼ 急求！！！用C語言編寫一個編譯原理實驗的簡單優先分析法程序
編譯原理IF條件語句的翻譯程序設計—簡單優先法、輸出四元式通過設計、編制、調試一個條件語句的語法及語義分析程序，加深對語法及語義分析原理的理解，並實現詞法分析程序對單詞序列的詞法檢查和分析。具體做到以下幾點：①對輸入語句進行詞法分析。將輸入的字元串進行掃描和分解，識別出一個個合法的單詞。單詞種類包括：關鍵字，標識符，運算符，常數和界限符②進行語法分析。編寫條件語句的相應文法，按照語法分析方法中的簡單優先分析法為文法設計簡單優先表，對詞法分析得到的單詞序列進行語法分析，以判別輸入的語句是否屬於該文法的條件語句。③語法制導翻譯。設計中間代碼（四元式）序列的結構及屬性文法，運用語法制導翻譯，在進行語法分析的同時，執行相應的語義規則描述的動作，從而實現語義處理，生成中間代碼以四元式的形式輸出。④錯誤提示。對不同的錯誤給出簡略描述，並終止程序的繼續執行。下載地址如下，有你要的東西！pile.rar
❽ 編譯原理詞法分析器是干什麼用的，怎麼用
1、識別出源程序中的各個單詞符號，並轉換成內部編碼形式 2、刪除無用的空白字元回車字元以及其他非實質性字元 3、刪除注釋 4、進行詞法檢查，報告所發現的錯誤。
❾ 編譯原理 詞法分析
C語言詞法分析器
#include<iostream>
#include<stdio.h>
#include<string>
using namespace std;
FILE *f; //定義一個文件變數
static int line = 1; //表示游標所在的行數
struct ID{ char *name; int count;}id[100];//用於存放ID號碼
static int I = 0; //用於記錄ID存放的數量
int Number[100]; //用於存放數字
static int P = 0; //用於記錄存放數字的個數
int error[100] = {0}; //用於記錄錯誤所在的行數
static int K = 0; //記錄錯誤次數
void Error(); //記錄錯誤
void loginID(char *); //注冊ID號
void loginNumber(int &); //記錄數字
void noteLine(char &); //記錄游標所在的行數
void print(); //輸出分析結果
int same(char *chr); //判斷單詞是否已經存在
void Error()
{ error[K++] = line; }
void loginID(char *chr) //注冊ID號
{
int k = 0;
int h = 0;
for(int i = 0; i < I; i++)
{
if(!strcmp(chr,id.name)) //如果單詞已經存在
{
id.count++;
k = 1;
}
}
if(k == 0) //該單詞不存在
{
h = I + 1;
//I = h;
id[h].count++;
id[h].name = chr;
//strcpy(id[h].name ,chr);
}
}
void loginNumber(int &nu)
{ Number[P++] = nu; }
void noteLine(char &ch)
{
if ( ch == ' ' )
++line;
}
void print()//輸出部分
{
//cout << "關鍵字以及變數:" << endl;
//for(int i = 0; i < 100; i++)
//cout << i <<" " << id.name << " " << id.count << endl;
cout << "數字：" << endl;
for(int i = 1; i <= P; i++)
cout << i << ": " << Number[i-1] << endl;
if(error[0] != 0)
{
cout << "出現的錯誤！" << endl;
for(int i = 1; i <= K; i++)
cout << "第" << i << "個錯誤： " << "第" << error[i-1] << "行" << endl;
}
else cout << "沒有錯誤！" << endl;
}
//文件處理部分
void noblank( char &ch) //跳過空格，回車
{
noteLine(ch);
while(ch == ' ' || ch == ' ')
ch = fgetc(f);
}
void identifier(char name[],char &ch)//字母變數
{
int i;
for(i = 0; i < 20; i++)
name = '';
i = 0;
while (('0'<= ch && ch <= '9')||('a'<= ch&&ch <= 'z')||('A'<= ch&&ch <='Z'))
{
name = ch;
i++;
ch = fgetc(f);
}
loginID(name);
//for(int j = 0; j < i; j++)
//{cout << name[j];}
// cout << ' ';
}
int number(char &ch)//數字
{
int num=0;
while('0'<= ch && ch <= '9')
{
num = num* 10 + (ch-'0');
ch = fgetc(f);
}
if( ('a'<= ch&&ch <= 'z')||('A'<= ch&&ch <='Z'))
{
Error();
}
else if( ch == '.')
{;}
loginNumber(num); //記錄數字
return num;
}
void test(char &ch)//符號
{
char str[2]={'0/'};
if(ch == '*')
{ str[0] = ch; ch = fgetc(f);}
if(ch == '.')
{ str[0] = ch; ch = fgetc(f);}
if(ch == ',')
{ str[0] = ch; ch = fgetc(f);}
if(ch == '"')
{ str[0] = ch; ch = fgetc(f);}
if(ch == '/')
{ str[0] = ch; ch = fgetc(f);}
if(ch == '%')
{ str[0] = ch; ch = fgetc(f);}
if(ch == '^')
{ str[0] = ch; ch = fgetc(f);}
if(ch == '-')
{ str[0] = ch; ch = fgetc(f);}
if(ch == '{')
{ str[0] = ch; ch = fgetc(f);}
if(ch == '}')
{ str[0] = ch; ch = fgetc(f);}
if(ch == '[')
{ str[0] = ch; ch = fgetc(f);}
if(ch == ']')
{ str[0] = ch; ch = fgetc(f);}
if(ch == ';')
{str[0] = ch; ch = fgetc(f);}
if(ch == ':')
{ str[0] = ch; ch = fgetc(f);}
if(ch == '?')
{ str[0] = ch; ch = fgetc(f);}
if(ch == '(')
{ str[0] = ch; ch = fgetc(f);}
if(ch == ')')
{str[0] = ch; ch = fgetc(f);}
if(ch =='+')
{
str[0] = ch;
if((ch = fgetc(f)) == '+' )
{
str[1] = ch;
ch = fgetc(f);
//cout << str[0] << str[1] << endl;
}
//cout << str[0]<< endl;
}
if(ch == '-')
{
str[0] = ch;
if((ch = fgetc(f)) == '-' )
{
str[1] = ch;
ch = fgetc(f);
//cout << str[0] << str[1] << endl;
}
//cout << str[0]<< endl;
}
if(ch == '&')
{
str[0] = ch;
if((ch = fgetc(f)) == '&' )
{
str[1] = ch;
ch = fgetc(f);
//cout << str[0] << str[1] << endl;
}
//cout << str[0]<< endl;
}
if(ch == '|')
{
str[0] = ch;
if((ch = fgetc(f)) == '|' )
{
str[1] = ch;
ch = fgetc(f);
//cout << str[0] << str[1] << endl;
}
//cout << str[0]<< endl;
}
if(ch == '!')
{
str[0] = ch;
if((ch = fgetc(f)) == '=' )
{
str[1] = ch;
ch = fgetc(f);
//cout << str[0] << str[1] << endl;
}
//cout << str[0]<< endl;
}
if(ch == '=')
{
str[0] = ch;
if((ch = fgetc(f)) == '=' )
{
str[1] = ch;
ch = fgetc(f);
//cout << str[0] << str[1] << endl;
}
}
if(ch == '>')
{
str[0] = ch;
if((ch = fgetc(f)) == '=' )
{
str[1] = ch;
ch = fgetc(f);
//cout << str[0] << str[1] << endl;
}
else
if(ch == '>' )
{
str[1] = ch;
ch = fgetc(f);
//cout << str[0] << str[1] << endl;
}
}
if(ch == '<')
{
str[0] = ch;
if((ch = fgetc(f)) == '=' )
{
str[1] = ch;
ch = fgetc(f);
}
else
if(ch == '<' )
{
str[1] = ch;
ch = fgetc(f);
}
}
}
int main()
{
char ch;
char name[30];
for(int i = 0; i < 30; i++)
name = '/0';
f = fopen("c.txt","r"); //打開指定輸入文件
if (f == NULL)
cout<<"文件不存在！"<<endl;
ch = fgetc(f);
while(!feof(f))
{
noblank( ch ); //跳過回車，空格
if( ( ch >= 'a' && ch <= 'z' )||( ch >= 'A' && ch <= 'Z' ))
{ identifier(name,ch); } //處理字母
else if( ch >= '0'&& ch <= '9')
{ number(ch); } //處理數字
else
{ test(ch); } //處理符號
}
print(); //列印詞法分析結果
fclose(f); //關閉文件
system("pause");
return 0;
}
❿ 編譯原理 詞法分析 要求輸入一個源文件，或是text形式的，然後對該文件進行詞法分析。要簡單一點的。
#include <iostream>
#include <vector>
#include <string>
#include <fstream>
using namespace std;
/*用來存儲目標文件名*/
string file_name;
/*提取文本文件中的信息。*/
string GetText();
/*獲得一個單詞符號,從位置i開始查找。
//並且有一個引用參數j，用來返回這個單詞最後一個字元在str的位置。*/
string GetWord(string str,int i,int& j);
/*這個函數用來除去字元串中連續的空格和換行
//第一個參數為目標字元串，第二個參數為開始位置
//返回值為連續的空格和換行後的第一個有效字元在字元串的位置*/
int DeleteNull(string str,int i);
/*判斷i當前所指的字元是否為一個分界符，是的話返回真，反之假*/
bool IsBoundary(string str,int i);
/*判斷i當前所指的字元是否為一個運算符，是的話返回真，反之假*/
bool IsOperation(string str,int i);
/*此函數將一個pair數組輸出到一個文件中*/
void OutFile(vector<pair<int,string> > v);
/*此函數接受一個字元串數組，對它進行詞法分析，返回一個pair型數組*/
vector<pair<int,string> > analyst(vector<string> vec);
/*此函數判斷傳遞的參數是否為關鍵字，是的話，返回真，反之返回假*/
bool IsKey(string str);
int main()
{
cout<<"*****************************\n";
cout<<"\n\nright: Archerzei\n\n\n";
cout<<"*****************************\n\n";
string com1=" ";
string com2="\n";
string fileline=GetText();
int begin=0,end=0;
vector<string> array;
do
{
begin=DeleteNull(fileline,begin);
string nowString;
nowString=GetWord(fileline,begin,end);
if(end==-1)
break;
if(nowString.compare(com1)&&nowString.compare(com2))
array.push_back(nowString);
begin=end+1;
}while(true);
vector<pair<int,string> > mid_result;
mid_result=analyst(array);
OutFile(mid_result);
cout<<"**********************************************************************\n";
cout<<"***程序已完成詞法分析，分析結果已經存儲在文件"<<file_name<<"中！！！***\n";
cout<<"**********************************************************************\n";
system("pause");
return 0;
}
/*提取文本文件中的信息*/
string GetText()
{
string file_name1;
cout<<"請輸入源文件名（包括路徑和後綴名）：";
cin>>file_name1;
ifstream infile(file_name1.c_str(),ios::in);
if (!infile)
{
cerr<<"無法打開文件! "<<file_name1.c_str()<<" !!!"<<endl;
exit(-1);
}
cout<<endl;
char f[1000];
infile.getline(f,1000,EOF);
infile.close();
return f;
}
/*獲得一個單詞符號,從位置i開始查找。
//並且有一個引用參數j，用來返回這個單詞最後一個字元在原字元串的位置。*/
string GetWord(string str,int i,int& j)
{
string no_use("(){} , ; \n+=*/-<>\"");
j=str.find_first_of(no_use,i);
if(j==-1)
return "";
if(i!=j)
j--;
return str.substr(i,j-i+1);
}
/*這個函數用來除去字元串中連續的空格和換行
//第一個參數為目標字元串，第二個參數為開始位置
//返回值為連續的空格和換行後的第一個有效字元在字元串的位置*/
int DeleteNull(string str,int i)
{
for(;;i++)
if(str[i]!=' '&&str[i]!='\n')
return i;
}
/*判斷i當前所指的字元是否為一個分界符，是的話返回真，反之假*/
bool IsBoundary(string str,int i)
{
int t;
char arr[7]={',',';','{','}','(',')','\"'};
for (t=0;t<7;t++)
if(str[i]==arr[t])
return true;
return false;
}
/*判斷i當前所指的字元是否為一個運算符，是的話返回真，反之假*/
bool IsOperation(string str,int i)
{
int t;
char arr[7]={'+','-','*','/','=','<','>'};
for (t=0;t<7;t++)
if(str[i]==arr[t])
return true;
return false;
}
/*此函數將一個個字元串數組輸出到一個文件中*/
void OutFile(vector<pair<int,string> > v)
{
cout<<"請輸入目標文件名（包括路徑和後綴名）：";
cin>>file_name;
ofstream outfile(file_name.c_str(),ios::out);
if (!outfile)
{
cerr<<"無法打開文件! "<<file_name.c_str()<<" !!!"<<endl;
exit(-1);
}
cout<<endl;
int i;
cout<<"*****************************\n";
cout<<"\n\nright: Archerzei\n\n\n";
cout<<"*****************************\n\n";
for(i=0;i<v.size();i++)
outfile<<"<"<<v[i].first<<" , \""<<v[i].second<<"\">"<<endl;
outfile<<"\n\n*********************************\n";
outfile.close();
return;
}
/*此函數接受一個字元串數組，對它進行詞法分析，返回一個pair型數組*/
vector<pair<int,string> > analyst(vector<string> vec)
{
vector<pair<int,string> > temp;
int i;
for(i=0;i<vec.size();i++)
{
if(vec[i].size()==1)
{
if((vec[i]==">"||vec[i]=="<"||vec[i]=="!")&&vec[i+1]=="=")
{
string jk=vec[i];
jk.append(vec[++i],0,1);
pair<int,string> pp(4,jk);
temp.push_back(pp);
continue;
}
if((vec[i]=="+"&&vec[i+1]=="+")||(vec[i]=="-"&&vec[i+1]=="-"))
{
string jk=vec[i];
jk.append(vec[++i],0,1);
pair<int,string> pp(4,jk);
temp.push_back(pp);
continue;
}
if(IsBoundary(vec[i],0))
{
pair<int,string> pp(5,vec[i]);
temp.push_back(pp);
}
else if(IsOperation(vec[i],0))
{
pair<int,string> pp(4,vec[i]);
temp.push_back(pp);
}
else if(vec[i][0]<='9'&&vec[i][0]>='0')
{
pair<int,string> pp(3,vec[i]);
temp.push_back(pp);
}
else
{
pair<int,string> pp(2,vec[i]);
temp.push_back(pp);
}
}
else if(vec[i][0]<='9'&&vec[i][0]>='0')
{
pair<int,string> pp(3,vec[i]);
temp.push_back(pp);
}
else if(IsKey(vec[i]))
{
pair<int,string> pp(1,vec[i]);
temp.push_back(pp);
}
else
{
pair<int,string> pp(2,vec[i]);
temp.push_back(pp);
}
}
return temp;
}
/*此函數判斷傳遞的參數是否為關鍵字，是的話，返回真，反之返回假*/
bool IsKey(string str)
{
string p[16]={"char","double","int","long","double","float","for","while","do","break","continue","switch","short","case","return","if"};
vector<string> ppp(p,p+16);
int u;
for(u=0;u<ppp.size();u++)
if(!str.compare(ppp[u]))
return true;
return false;
}
/*finished*/
已經驗收過了，在VC6.0上運行沒有問題。程序很容易看懂的，報告的話自己寫寫就可以了。要是有分就好了…………哈哈！！！