使用递归评估 boolean 表达式

Question

I've wrote a program to recursively evaluate boolean expression. But when I run with input "(1|1)" , in the second recursion of function S, specifically in line result = result || P() result = result || P() instead of step into P() , it passes the line and returns result. However it works fine with input "(~1|1)" . How can I fix this?

#include <stdbool.h>
#include <stdlib.h>

char *s;
char *f;
bool P(), A();

// The grammar is:
// S -> P||S | P&&S | P
// P -> A | ~A
// A -> (S) | 0 | 1

bool S() {
    bool result = P(); 
    if (*s == '|') {
        s++; 
        result = result || P(); 
    } 
    else
        if (*s == '&') {
            s++; 
            result = result && P(); 
        } 
    return result;
}

bool P() {
    if (*s == '~') {
        s++;
        return !A();
    }
    else
    {
        return A(); 
    }
}

bool A() {
    bool result;
    if (*s == '(') {
        s++;
        result = S();
        if (*s == ')') s++;
        else {
            printf("Syntaktisch falsch! Stelle %ld, Art: Anzahl von '(' und ')' sind nicht gleich\n", s-f);
            exit(1);
        }
    } 
    else
        if (*s == '1') {
            s++;
            result = true; 
        }
        else
            if (*s == '0') {
                s++;
                result = false; 
            }
            else {
                printf("Syntaktisch falsch! Stelle: %ld, Art: Boolscher Ausdruck fehlt\n", s-f); // Hier gibt Fehlermeldung aus, wenn Input String ist nur '~'. 
                exit(1);
            }

    return result;
}

int main(int argc, char *argv[])
{
    bool result;
    if (argc != 2) exit(1);
    s = argv[1];
    f = s;
    result = S();
    printf("%d\n", result);
    return 0;
}

Answer 1

result || P() result || P() only requires that ONE of the values - result OR P() - be TRUE in order to satisfy the expression. If result is TRUE when result || P() result || P() is evaluated the function is not required to be called - this is known as "short-circuit evaluation". Change your code to:

bool p_val;

p_val = P();
result = result || p_val;

in order to ensure that the function is actually invoked.

Answer 2

result = result || P();

P will never will called if result is the true

result = result && P();

P will never will called if result is false

so place the call first to make sure it is called.

P() || result
P() && result

it is called https://en.wikipedia.org/wiki/Short-circuit_evaluation