C语言算术表达式语法分析程序：消除左递归并实现识别

#include <stdio.h> #include <stdlib.h> #include <string.h> #include using namespace std; char str[10]; int index = 0; void E(); // E->TX; void X(); // X->+TX|-TX|^ void T(); // T->FY void Y(); // Y->*FY|/FY|^ void F(); // F->(E) | i int count; int m, n; int len;

void rank1() { printf("%d ", count); count++; }

void analyze() { if (m < 0) cout << ' '; else { for (int i = 0; i <= m; i++) cout << str[i]; } cout << ' ' << ' ' << ' '; }

void latter() { printf("%c", str[index]); cout << ' ' << ' ' << ' '; }

void remain() { for (int j = n + 1; j <= len; j++) cout << str[j]; cout << '\n'; }

/* 文法G[E]: E → E+T | E-T | T T → T*F | T/F | F F → (E) | i 消除左递归后： E → TX X → +TX|-TX|^ T → FY Y → *FY | /FY|^ F → (E) | i */

int main() { count = 0; m = -1, n = -1; index = 0; printf("请输入算数表达式："); scanf("%s", str); len = strlen(str); str[len] = '#'; printf("步骤 文法 分析串 分析字符 剩余串\n"); E(); if (str[index] == '#') printf("是\n"); else printf("否\n"); return 0; }

void E() { rank1(); printf("E -> TX "); analyze(); latter(); remain(); T(); X(); }

void X() { if (str[index] == '+') { index++; m++; n++; rank1(); printf("X ->+TX "); analyze(); latter(); remain(); T(); X(); } else if (str[index] == '-') { index++; m++; n++; rank1(); printf("X ->-TX "); analyze(); latter(); remain(); T(); X(); } else { rank1(); printf("X -> ^ "); analyze(); latter(); remain(); } } void T() { rank1(); printf("T -> FY "); analyze(); latter(); remain(); F(); Y(); } void Y() { if (str[index] == '*') { index++; m++; n++; rank1(); printf("Y ->*FY "); analyze(); latter(); remain(); F(); Y(); } else if (str[index] == '/') { index++; m++; n++; rank1(); printf("Y ->/FY "); analyze(); latter(); remain(); F(); Y(); } else { rank1(); printf("Y -> ^ "); analyze(); latter(); remain(); } } void F() { if (str[index] == 'i') { index++; m++; n++; rank1(); printf("F ->i "); analyze(); latter(); remain(); } else if (str[index] == '(') { index++; m++; n++; rank1(); printf("F ->(E) "); analyze(); latter(); remain(); E(); if (str[index] == ')') { index++; m++; n++; rank1(); printf("F ->(E) "); analyze(); latter(); remain(); } else { printf("否\n"); exit(0); } } else { printf("否\n"); exit(0); }