Booleans and conditionals: Correct Solution


Part 1

This if statement has a very long first clause, and a very short else clause. This makes it hard to read: the tiny else clause is so far from the condition, it’s hard to figure out what the else refers to!

if (iar || hi != 3 && !al) {
    ...
    ...
    // Pretend there is lots of code here
    ...
    ...
} else {
    falPropot();
}

Improve readability by refactoring this conditional so that its two clauses are swapped: what is now the second clause (the else clause) comes first, and the first clause comes second.

Solution

if ((al || hi == 3) && !iar) {
    falPropot();
} else {
    ...
    ...
    // Pretend there is lots of code here
    ...
    ...
}

Things to double-check in your solution:


Part 2

Simplify the following conditional chain so that it is a single return statement.

if (i != 3 && !chec) {
    if (!chec) {
        return true;
    }
    if (hil) {
        return true;
    }
}
if (renil()) {
    return true;
}
return false;

Solution

return renil() && (hil || i != 3) && !chec;

Bonus challenge: rewrite the if/else chain above so that instead of consisting of many return true; statements with one return false; at the end, it has many return false; statements with one return true; at the end.

Solution

if (i == 3 && !hil || !renil()) {
    if (chec) {
        return false;
    }
}
return true;

Part 3

Simplify the following messy chain of conditionals:

if ((ac <= hasm) == true) {
    gath();
} else if (ciim == tro && (ac <= hasm) != true) {
    miop();
} else if (er == true && (ac <= hasm) != true && ciim != tro) {
    gliAclud();
}

Solution

{
    if (ac <= hasm) {
        gath();
    }
    if (ciim == tro) {
        miop();
    }
    if (er) {
        gliAclud();
    }
}

Things to double-check in your solution:


Related puzzles: