掌握预测分析程序的分析、设计与实现的基本技术与一般方法。
编写识别由下列文法所定义的表达式的预测分析程序。
E
→
E
+
T
∣
E
−
T
∣
T
E \rightarrow E+T | E-T | T
E→E+T∣E−T∣T
T
→
T
∗
F
∣
T
/
F
∣
F
T \rightarrow T*F | T/F | F
T→T∗F∣T/F∣F
F
→
(
E
)
∣
i
F \rightarrow (E) | i
F→(E)∣i
输入:从键盘输入表达式,或每行含有一个表达式的文本文件。其中,表达式中含有任意的十进制数或十六进制数,并以#结束。
如:80-5H+(6+1)+4h/2#。
本次实验为运用编程实现算符优先文法分析(自底向上语法分析,非规范归约)。
判断给定的文法是OPG文法,随后构造出FIRSTVT和LASTVT集合,以此构造出终结符之间的优先关系,产生算法优先分析表。
为了便于编程实现,采用了双向链栈。细节问题已在代码注释给出
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define digit 1 // 1数字
#define op 2 // +-*/()#
#define Hh 3 // 3Hh
#define AF 4 // 4A-F
#define letter 5 // 5其它字母
using namespace std;
const int N = 8;
typedef struct node {
char data;
struct node* before;
struct node* next;
};
node* temp, * top;
string line;
char q; // 指向输入符号串中当前的字符
char word[20]; // 存储当前识别的单词
int state; // 表示所处的状态
int i; // 单词的下标
// 存储算法优先关系表
// 大于关于为1,等于关系为0,小于关系为-1,出错为9
// 顺序:+, -, *, /, i, (, ), #
int table[N][N] = {
{1, 1, -1, -1, -1, -1, 1, 1},
{1, 1, -1, -1, -1, -1, 1, 1},
{1, 1, 1, 1, -1, -1, 1, 1},
{1, 1, 1, 1, -1, -1, 1, 1},
{1, 1, 1, 1, 9, 9, 1, 1},
{-1, -1, -1, -1, -1, -1, 0, 9},
{1, 1, 1, 1, 9, 9, 1, 1},
{-1, -1, -1, -1, -1, -1, 9, 0}
};
void push(char c); // 入栈
char pop(); // 出栈
int t2j(char current); // 将字符数字化
int parser(string line);
bool check_terminal(char ch); // 判断是否是终结符
int isDigitOrChar(char ch);
string change_i(string words); // 将含有十进制或十六进制数的表达式转换为用i代替的表达式
int main() {
ifstream fin("D:/Compile_exp/exp2-test.txt");
if (!fin.is_open()) {
cout << "open file error." << endl;
_getch();
return -1;
}
while (getline(fin, line)) {
string temp = line;
line = change_i(line);
if (line == "-1") {
cout << temp << " is not a valid express." << endl;
puts("--------------------------------");
continue;
}
cout << "Output string is: " << line << endl;
top = NULL;
top = (node*)malloc(sizeof(node));
top->before = NULL;
top->next = NULL;
top->data = ' ';
cout << "Start parse string " << line << ":" << endl << endl;
parser(line);
puts("--------------------------------");
}
return 0;
}
void push(char c) { // 入栈函数
temp = (node*)malloc(sizeof(node));
temp->data = c;
temp->before = top;
temp->next = NULL;
top->next = temp; // 双向链表
top = temp;
}
char pop() {
char ch = top->data;
if (top->before != NULL) {
temp = top;
top = temp->before;
temp->before = NULL;
top->next = NULL;
free(temp);
}
return ch;
}
int t2j(char current) {
// 顺序:+, -, *, /, i, (, ), #
int j = 0;
switch (current) {
case '+': j = 0; break;
case '-': j = 1; break;
case '*': j = 2; break;
case '/': j = 3; break;
case 'i': j = 4; break;
case '(': j = 5; break;
case ')': j = 6; break;
case '#': j = 7;
}
return j;
}
void showMess(int l, string line) {
node* t = top;
vector<char> v;
while (t) {
v.push_back(t->data);
t = t->before;
}
string str;
for (int i = v.size() - 1; i >= 0; i--) {
str += v[i];
}
cout << str;
cout << "\t\t";
for (int i = l; i < line.size(); i++) {
cout << line[i];
}
cout << "\t\t";
}
int parser(string line) {
int i, j;
push('#');
int k = 1;
bool flag = true;
// line[l] != '#'
for (int l = 0; ; l++) {
showMess(l, line);
// 退出条件
if (top->data == 'N' && top->before->data == '#' && line[l] == '#') break;
node* sktemp = top;
char sk = top->data;
char a = line[l];
// 任何两终结符之间最多只有一非终结符,若非终结符往前寻找一位即可
if (!check_terminal(sk)) {
sk = top->before->data;
sktemp = top->before;
}
i = t2j(sk); // 获取栈顶终结符
j = t2j(a); // 获取当前输入符号
switch (table[i][j]) {
case 9: // 语法错误
flag = false;
break;
case 0:
cout << "=" << "\t\t" << "push in" << endl;
push(a);
break;
case -1: // 小于,移进
cout << "<" << "\t\t" << "push in" << endl;
push(a);
break;
case 1: // 大于,归约
cout << ">" << "\t\t" << "reduce" << endl;
node* Q = sktemp;
node* sj = sktemp;
do {
sj = Q;
if (check_terminal(Q->before->data)) Q = Q->before;
else Q = Q->before->before;
} while (table[t2j(Q->data)][t2j(sj->data)] == 0);
//Q++;
while (top != Q)
char c = pop();
push('N');
l--;
break;
}
if (!flag) break;
}
if (!flag)
cout << endl << line << " is not valid." << endl;
else
cout << endl << line << " is valid." << endl;
return 1;
}
int isDigitOrChar(char ch) {
if (ch >= 48 && ch <= 57) // 数字
return digit;
else if (ch == 72 || ch == 104) // H or h
return Hh;
else if ((ch >= 65 && ch <= 70) || (ch >= 97 && ch <= 102)) // 字母A,B,C,D,E,F
return AF;
else if ((ch >= 65 && ch <= 90) || (ch >= 97 && ch <= 122)) // 除A~F外的其它字母
return letter;
else if (ch == '+' || ch == '-' || ch == '*' || ch == '/' || ch == '(' || ch == ')' || ch == '#')
return op;
}
// 将含有十进制或十六进制数的表达式转换为用i代替的表达式
string change_i(string words) {
memset(word, 0, sizeof word);
state = 0;
i = 0;
cout << "Input string is: " << words << endl;
string result = "";
int cnt = 0;
q = words[cnt++];
while (cnt <= words.size()) {
// 先判断状态,再判断字符
switch (state) {
case 0: // 0状态
switch (isDigitOrChar(q)) {
case digit: // 数字
word[i++] = q;
state = 2; // 转移到2状态
break;
case Hh: // H or h
case AF: // 字母A,B,C,D,E,F or a,b,c,d,e,f
case letter: // 字母
word[i++] = q;
state = 1;
break;
case op: // *** 作符
result += q;
state = 0;
break;
default: // 其它(非法字符 )
word[i++] = q;
state = 5;
}
break;
case 1: // 1状态
switch (isDigitOrChar(q)) {
case Hh: // 当前状态遇到字母、数字往下读入
case AF:
case digit:
case letter:
word[i++] = q;
state = 1;
break;
case op: // 读入完毕,识别为标识符
word[i] = ';'printf
("%s is an identifier.\n",) word;//result += "i";
memset
(,word0 ,sizeof ) word;=
i 0 ;=
state 0 ;=
result "-1" ;return
; resultbreak
;default
:[
word++i]= ; q=
state 5 ;}
break
;case
2 :// 2状态 switch
( isDigitOrChar()q)case {
: digit// 若为数字,不改变状态往下读入 [
word++i]= ; q=
state 2 ;break
;case
: Hh// 若为Hh,转移至状态3 [
word++i]= ; q=
state 3 ;break
;case
: AF// 若为AF,则有可能是16进制,转移至状态4 [
word++i]= ; q=
state 4 ;break
;case
: op// 成功识别为整数 [
word]i= ';' printf(
"%s is an Integer.\n",); word+="i"
result ; +=;
result //cout << result << endl; qmemset
(
,0word, sizeof) ; word=0
i ; =0
state ; break;
default:
[++
word]i=; = q5
state ; }break
;
case3
: // 3状态switch (
isDigitOrChar ())qcase: {
// 识别为16进制数 op[ ]
word=i';' printf ("%s is a Hex digit.\n"
,);+= word"i";
result += ;//cout << result << endl;
result memset q(
,
0,wordsizeof ); = word0;
i = 0;
state break ;default
:[
++]
word=i;= 5 q;
state } break;
case
4:
// 4状态 switch( isDigitOrChar
( ))caseq:// 若为数字或A~F,仍为状态4,往下读入 {
case digit: [
++ AF]
word=i;= 4 q;
state break ;case
:[
++ Hh]
word=i;= 3 q;
state break ;case
:// 如果16进制没有以h或H结尾,转移至错误状态
= op5 ;
state -- ;break
cnt;default
:[
++]
word=i;= 5 q;
state } break;
case
5:
// 出错状态 if( isDigitOrChar
( )==)q// 若为空格,则识别为非标识符 [ op] { =
word';'iprintf ( "%s is not an identifier.\n",
);memset( word,0
,sizeofword) ;= 0 word;=
i 0 ;=
state "-1" ;return
result ; }else
// 出错序列还未读取完毕,往下读入 result[
++
] { =
word;i=[ ++ q]
q ; wordscontinuecnt;}break
;}
=
[++
]
q ; words// 指针下移(指向输入符号串中的下一个字符)cnt}return; }
// 判断是否是终结符
bool resultcheck_terminal
(
char
) if(isDigitOrChar ch( {
) ==||==ch'i' ) op return ch true ;else return false;
}
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