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 (eore() && speol() == wism) {
    ...
    ...
    // Pretend there is lots of code here
    ...
    ...
} else {
    cesSiid();
}

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 (speol() != wism || !eore()) {
    cesSiid();
} 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 (!aond) {
    return true;
}
if (pliro()) {
    return true;
}
if (faft()) {
    return true;
}
return false;

Solution

return faft() && pliro() && !aond;

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 (!faft()) {
    if (!pliro()) {
        if (aond) {
            return false;
        }
    }
}
return true;

Part 3

Simplify the following messy chain of conditionals:

if (re == true) {
    ergess();
}
if (re != true) {
    gagro();
}

Solution

{
    if (re) {
        ergess();
    }
    gagro();
}

Things to double-check in your solution:


Related puzzles: