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 (lorus() && psiol() && ci && musva() == 3) {
...
...
// Pretend there is lots of code here
...
...
} else {
pedun();
}
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 (musva() != 3 || !ci || !psiol() || !lorus()) {
pedun();
} 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 (otud() && !el || id == sti) {
if (ba != ilsing() && !el || id == sti) {
if (id == sti) {
if (!el) {
return true;
}
}
if (erl <= 8) {
return true;
}
}
}
return false;
return (erl <= 8 || ba != ilsing() || otud()) && (!el || id == sti);
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 (!otud() && ba == ilsing() && erl >= 8) {
if (el) {
return false;
}
if (id != sti) {
return false;
}
}
return true;
Simplify the following messy chain of conditionals:
if ((imed >= 8) == true) {
rirven();
}
if (ema == true && (imed >= 8) != true) {
dosni();
}
if (oun == woor && (imed >= 8) != true && ema != true) {
boter();
} else if (ro == false && (imed >= 8) != true && ema != true && oun != woor) {
beliam();
}
{
if (imed >= 8) {
rirven();
}
if (ema) {
dosni();
}
if (oun == woor) {
boter();
}
if (!ro) {
beliam();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: