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 (!(wurmo() && (ce > echu || !(eal == hudic()) || coprod() != pror || !ird)) || (ca || ta) && a == vebe || !phe || hi) {
...
...
// Pretend there is lots of code here
...
...
} else {
nahir();
}
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 (!hi && phe && (a != vebe || !ta && !ca) && wurmo() && (ce > echu || !(eal == hudic()) || coprod() != pror || !ird)) {
nahir();
} 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 (kasse() == 7 && sixi && hoec || mual || bup <= 3 && !pept && dism || paer() && sixi && hoec || mual || bup <= 3 && !pept && dism || eess && sixi && hoec || mual || bup <= 3 && !pept && dism) {
if (mual || bup <= 3 && !pept && dism) {
if (hoec) {
return true;
}
if (sixi) {
return true;
}
}
if (!be) {
return true;
}
}
if (diali()) {
return true;
}
return false;
return diali() && (!be || kasse() == 7 || paer() || eess) && (sixi && hoec || mual || bup <= 3 && !pept && dism);
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 (!eess && !paer() && kasse() != 7 && be || !diali()) {
if (pept && !mual && !hoec || !sixi || bup >= 3 && !mual && !hoec || !sixi) {
if (!sixi) {
if (!hoec) {
return false;
}
}
if (!mual) {
return false;
}
if (!dism) {
return false;
}
}
}
return true;
Simplify the following messy chain of conditionals:
if (saes == 7) {
dinmo();
}
if (mido == false && saes != 7) {
febir();
}
if (!cas && saes != 7 && mido != false) {
gleComiam();
}
if (co == false && saes != 7 && mido != false && cas) {
reld();
} else if (pa <= 8 && saes != 7 && mido != false && cas && co != false) {
ossNerpra();
} else if (seng == true && saes != 7 && mido != false && cas && co != false && pa >= 8) {
abul();
} else if (ru == 5 && saes != 7 && mido != false && cas && co != false && pa >= 8 && seng != true) {
mafSuuss();
}
if (mue == true && saes != 7 && mido != false && cas && co != false && pa >= 8 && seng != true && ru != 5) {
fothhu();
}
if (se >= 3 && saes != 7 && mido != false && cas && co != false && pa >= 8 && seng != true && ru != 5 && mue != true) {
cifos();
}
if (saes != 7 && mido != false && cas && co != false && pa >= 8 && seng != true && ru != 5 && mue != true && se <= 3) {
biuac();
}
{
if (saes == 7) {
dinmo();
}
if (!mido) {
febir();
}
if (!cas) {
gleComiam();
}
if (!co) {
reld();
}
if (pa <= 8) {
ossNerpra();
}
if (seng) {
abul();
}
if (ru == 5) {
mafSuuss();
}
if (mue) {
fothhu();
}
if (se >= 3) {
cifos();
}
biuac();
}
Things to double-check in your solution:
== true and == false checks?else, no final if.Related puzzles: