我们先撕简单的。a ab a|b aa* a(a|b)* 先不管匹配任意字符的. 重复>=1次的+ [^0-9]除0-9外 \digit数字等。
正则表达式(regular expression, re)为啥叫表达式,不叫正则字符串之类?因为它是个表达式。3+5*2是个表达式;两个字符串可以有连接运算,如"a"+"b"或"a"."b"得到"ab"。
在正则表达式里,a,b,c就像2,3,5,是被运算的数,. | * ()是运算符。注意:ab是a . b,人们为了省事不把.写出来。
(3+5)*2=16,3+(5*2)=13。如果没有四则运算优先级,3+5*2等于16还是13?运算符后置(后缀表达式)没有歧义,例如35+2*是mul(add(3,5), 2),352+*是mul(3, add(5,2))。
What are infix, postfix and prefix expressions?
我们分3步走:
- 把a(a|b)*变成aab|*.这样的后缀表达式,40行程序
- 用Thompson算法把后缀表达式变成NFA,算14行程序
- 用NFA检查是否匹配,算11行程序
第一步中缀变后缀请看代码。
第二步后缀变NFA。NFA可以像积木一样拼起来。下面分别是a, a*, a|b的NFA:
图片是用dot - graphviz version 2.49.0画的。如 dot -o ab.png -Tpng todot.txt 或 dot -Tpng todot.txt >ab.png 。dot -h看帮助。
拼接NFA的代码:
NFA postfix_to_nfa(const char* pfstr) { Stack<NFA> stk; for (const char* p = pfstr; *p; p++) { switch (*p) { case '.': stk.push(stk.pop() + stk.pop()); break; case '|': stk.push(stk.pop() | stk.pop()); break; case '*': stk.push(*stk.pop()); break; default: stk.push(*p); } } NFA nfa = stk.pop(); if (!stk.empty()) error; return nfa; }
运算符函数也不长,全部代码180行:
// 从ChrisZZ([email protected])的程序改来的 #include <stdio.h> #include <string.h> #include <string> #include <stack> using namespace std; #define error throw __LINE__ template<class T>struct Stack : public stack<T> { T pop() { T t = top(); stack<T>::pop(); return t; } }; const char END = '\0', EPSILON = '\001'; // Epsilon (upper case Ε, lower case ε): empty struct State { // 像链表里的node int id; // 自动加1的编号 State* next[2]; // 到next[0]的边是epsilon;到next[1]的是char char ch; State(int ch_=256, State* p1=0, State* p0=0) : id(_id++), ch(ch_) { next[0] = p0; next[1] = p1; } static int _id; static char _visited[256]; // 下标是State的编号,仅print时用 }; int State::_id; char State::_visited[256]; struct NFA { State *start, *end; NFA() : start(0), end(0) {} NFA(char ch) { end = new State(END); start = new State(ch, end); } NFA operator + (NFA nfa) { end->ch = EPSILON; end->next[1] = nfa.start; end = nfa.end; return *this; } NFA operator | (NFA nfa) { State *head = new State(EPSILON, start, nfa.start), *tail = new State(END); end->ch = EPSILON; end->next[1] = tail; end = tail; start = head; nfa.end->ch = EPSILON; nfa.end->next[1] = tail; return *this; } NFA operator * () { State *tail = new State(END), *head = new State(EPSILON, start, tail); end->ch = EPSILON; end->next[0] = start; end->next[1] = tail; end = tail; start = head; return *this; } void print(const char* file_name); const char* elm; // point to the end of the longest match const char* match(const char* str) { elm = str; visit4m(start, str); return elm; } void visit4p(const State* s, FILE* fp); // visit for print void visit4m(const State* s, const char* str); // visit for match }; NFA postfix_to_nfa(const char* pfstr) { Stack<NFA> stk; for (const char* p = pfstr; *p; p++) { switch (*p) { case '.': stk.push(stk.pop() + stk.pop()); break; case '|': stk.push(stk.pop() | stk.pop()); break; case '*': stk.push(*stk.pop()); break; default: stk.push(*p); } } NFA nfa = stk.pop(); if (!stk.empty()) error; return nfa; } void NFA::print(const char* file_name) { // 同时输出到屏幕和DOT文件 puts(""); FILE* fp = fopen(file_name, "wt"); if (!fp) return; fputs("digraph {\n\"\"\n", fp); fputs("[shape = plaintext]\n", fp); fputs("\trankdir = LR\n", fp); memset(State::_visited, 0, sizeof(State::_visited)), visit4p(start, fp); fputs("}", fp), fclose(fp); } void NFA::visit4p(const State* st, FILE* fp) { if (State::_visited[st->id]) return; State::_visited[st->id] = 1; for (int i = 0; i < 2; i++) { if (State* p = st->next[i]) { char label[16]; if (st->ch == EPSILON) strcpy(label, "''"); else sprintf(label, "'%c'", st->ch); // DOT支持不带BOM的UTF-8编码的文件。ε的UTF-8编码是\xce\xb5 printf("%d - %s -> %d\n", st->id, label, p->id); fprintf(fp, "%d -> %d [label = <%s>]\n", st->id, p->id, label); visit4p(p, fp); } } } void NFA::visit4m(const State* st, const char* str) { if (st == end) { if (str > elm) elm = str; return; } for (int i = 0; i < 2; i++) { if (State* p = st->next[i]) { if (st->ch == EPSILON) visit4m(p, str); if (st->ch == *str) visit4m(p, str + 1); } } } struct CountOf { int opnd; // a是opnd b是opnd ab.也是opnd int or; // | }; string re_to_postfix(const char* re) { string out; CountOf cntof = { 0 }; stack<CountOf> khdz; // KuoHao (parenthesis) 的栈 const char* p; for (p = re; *p; p++) { switch (char c = *p) { case '(': if (cntof.opnd > 1) out += '.'; // a(??? khdz.push(cntof); cntof.or = cntof.opnd = 0; break; case ')': if (cntof.opnd == 0 || khdz.empty()) error; // ) () while (--cntof.opnd > 0) out += '.'; // ((a|b)(c|d)) =1时不进循环 1个opnd不需要. while (cntof.or-- > 0) out += '|'; // =1时进循环 cntof = khdz.top(); khdz.pop(); ++cntof.opnd; // 如遇到(时还没有opnd,遇到(a)的)时,知道了(a)是个opnd break; case '*': if (cntof.opnd ==0 ) error; out += c; break; case '|': // a|b变ab| a|b|c变ab|c| ab|c变ab.c| if (cntof.opnd == 0) error; while (--cntof.opnd > 0) out += '.'; ++cntof.or; break; default: // a变a ab变ab. abc变ab.c. if (cntof.opnd > 1) { --cntof.opnd; out += '.'; } out += c; ++cntof.opnd; } // switch // printf("%*c", 5, ' ')输出5个空格 printf("%*c%s %d %d %s\n", p - re, ' ', p, cntof.opnd, cntof.or, out.c_str()); } // for if (!khdz.empty()) error; while (--cntof.opnd > 0) out += '.'; while (cntof.or-- > 0) out += '|'; printf("%*c%s %s\n", p - re, ' ', p, out.c_str()); return out; } int main(){ try { //const char* re = "a"; //const char* re = "a*"; //const char* re = "ab"; //const char* re = "a|b"; const char* re = "((a|b)(c|d))*"; NFA nfa = postfix_to_nfa(re_to_postfix(re).c_str()); nfa.print("todot.txt"); const char* s = "bdabc"; const char* p = nfa.match(s); printf("\nmatch: %.*s\n", p - s, s); } catch(int n) { printf("Error at line %d.\n", n); } getchar(); return 0; }View Code
print和match都是递归遍历图。
todo
标签:const,NFA,cntof,正则表达式,stk,char,State From: https://www.cnblogs.com/funwithwords/p/16999293.html