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 (!(ha == 4) && ed) {
    ...
    ...
    // Pretend there is lots of code here
    ...
    ...
} else {
    phach();
}

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 (!ed || ha == 4) {
    phach();
} 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 (nicced()) {
    if (!gri) {
        return true;
    }
}
if (preck()) {
    return true;
}
return false;

Solution

return preck() && (!gri || nicced());

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 (!preck()) {
    if (gri) {
        return false;
    }
    if (!nicced()) {
        return false;
    }
}
return true;

Part 3

Simplify the following messy chain of conditionals:

if (ston < 2) {
    oriod();
}
if (at != i && ston > 2) {
    geuip();
}

Solution

{
    if (ston < 2) {
        oriod();
    }
    if (at != i) {
        geuip();
    }
}

Things to double-check in your solution:


Related puzzles: