c - C 中的递归下降解析器 - 跳过 epsilon 产生式

Question

构造一个递归下降解析器来解析以下语法。有没有一种方法可以让我返回而无需为所有 epsilon 作品(e)传递任何东西？ (考虑到我正在采用的解析方法，如代码所示)

E = TG
G = +TG | e
T = FH
H = *FH | e
F = (E) | id

#include <stdio.h>

char* next;

int terminal(char);
int E();int G();int G1();int G2();int T();int H();int H1();int H2();int F();int F1();int F2();

int main(int argc, char const *argv[])
{
    char str[10];
    printf("Enter an expression to be parsed : ");
    scanf("%s", str);
    next = &str[0]
    (*next == '\0' && E() == 1) ? printf("Parsed Successfully\n") : printf("Parsed Unsuccessfully\n");
    return 0;
}

int terminal(char token){return *next++ == token;}
int E(){return T() && G();}
int G(){char* temp = next; return (next = temp, G1()) || (next = temp, G2());}
int G1(){return terminal('+') && T() && G();}
int G2(){return;}   //ERROR : non-void function should return a value
int T(){return F() && H();}
int H(){char* temp = next; return (next = temp, H1()) || (next = temp, H2());}
int H1(){return terminal('*') && F() && H();}
int H2(){return;}   //ERROR : on-void function should return a value
int F(){char* temp = next; return (next = temp, F1()) || (next = temp, F2());}
int F1(){return terminal('(') && E() && terminal(')');}
int F2(){return terminal('a');}

Answer 1

代码不可读。但逻辑表明你应该返回一个 bool 值。空的生产总是成功的。所以它应该返回true。例如，

 int H2(){return true;} 

自然地，内联这些琐碎的产生式是有意义的。