Are there any problems with this Boolean based C programming question?

I’m currently learning about the Boolean data type, and I’m wondering if the conditions in the if-else statements are complete…

#include <stdio.h>
#include <stdbool.h>
int main() {
    bool a = false, b = false, c = false , d = false;
    if ((a&&b)||(!c&&d)) {
        if(((a||!b)&&c)||(b&&!a)) {
            printf("1");
        } else if ((a||(d&&b))&& !b) {
            printf("2");
        } else {
            printf("3");
        }
    } else {
        if(!(d&&c)&&(!a)) {
            printf("4");
        } else {
            printf("5");
        }
    }
    return 0;
}

Shouldn’t the conditions in the if-else statements have a comparison? For example (a&&b)==true || (!c&&d)==true. Running the program as it is outputs a "4" and I’m a bit confused as to why. Thank you

>Solution :

In order to understand how the code executes and gives the final result, the key is to understand the operator precedence.

Among the three logic operators, namely logic not !, logic and && ,and logic or ||, logic not ! has the highest precedence, the logic and && and then comes logic or ||.

The logic not ! has a higher precedence than logic and && and logic or ||, so the a&&b and !c&&d both evaluate to false.

Because (a&&b)||(!c&&d) evaluates to false, the outer if body doesn’t execute.

Now, it comes into the outer else section.

d&&c evaluates to false because both of d and c are false. As a result, !(d&&c) evaluates to true. Of course not a !a evaluates to true too.

In the end, !(d&&c)&&(!a) is true. So printf("4"); executes and output a 4