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 ((sloPrudi() || stel) && !eic) {
    ...
    ...
    // Pretend there is lots of code here
    ...
    ...
} else {
    gneAth();
}

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 (eic || !stel && !sloPrudi()) {
    gneAth();
} 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 (plor && !ner || vess() == enon && !ner) {
    if (!ner) {
        return true;
    }
    if (de != ecking()) {
        return true;
    }
}
return false;

Solution

return (de != ecking() || plor || vess() == enon) && !ner;

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 (vess() != enon && !plor && de == ecking()) {
    if (ner) {
        return false;
    }
}
return true;

Part 3

Simplify the following messy chain of conditionals:

if (va == true) {
    strecs();
} else if ((vedi != bir) == true && va != true) {
    istce();
} else if (clil == true && va != true && (vedi != bir) != true) {
    claun();
}

Solution

{
    if (va) {
        strecs();
    }
    if (vedi != bir) {
        istce();
    }
    if (clil) {
        claun();
    }
}

Things to double-check in your solution:


Related puzzles: