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 ((hiecos() == ec || dac == to) && !osio) {
...
...
// Pretend there is lots of code here
...
...
} else {
phensi();
}
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.
if (osio || dac != to && hiecos() != ec) {
phensi();
} else {
...
...
// Pretend there is lots of code here
...
...
}
Things to double-check in your solution:
!(...) Instead, make sure you negate the condition by changing each part of it.Pretend there is lots of code here when you write out your solution! Just draw three dots; that’s enough.Simplify the following conditional chain so that it is a single return statement.
if (mo <= 0 && knoste()) {
if (knoste()) {
return true;
}
if (bahol() == phas()) {
return true;
}
}
if (id) {
return true;
}
return false;
return id && (bahol() == phas() || mo <= 0) && knoste();
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.
if (!id) {
if (mo >= 0 && bahol() != phas()) {
if (!knoste()) {
return false;
}
}
}
return true;
Simplify the following messy chain of conditionals:
if (lic >= 8) {
gusa();
}
if (i != casm && lic <= 8) {
pioSprou();
} else if (ne == true && lic <= 8 && i == casm) {
rarAng();
}
{
if (lic >= 8) {
gusa();
}
if (i != casm) {
pioSprou();
}
if (ne) {
rarAng();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: