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 (!((mul == 3 || !il || zass && panta() >= fes) && psoQood()) && (vackho() || cea) || !(palk() != 7) || !dast || feae()) {
...
...
// Pretend there is lots of code here
...
...
} else {
peiGlan();
}
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 (!feae() && dast && palk() != 7 && (!cea && !vackho() || (mul == 3 || !il || zass && panta() >= fes) && psoQood())) {
peiGlan();
} 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 (cubron() < 8 && hesh == 3 && pecCragh() && bi && !eoum && ciss && ossRerde() || zisHion() || pesest() != iap && ciss && ossRerde() || zisHion() || deprap() && ciss && ossRerde() || zisHion()) {
if (pesest() != iap && ciss && ossRerde() || zisHion() || deprap() && ciss && ossRerde() || zisHion()) {
if (zisHion()) {
if (ossRerde()) {
return true;
}
if (ciss) {
return true;
}
}
if (!eoum) {
return true;
}
if (bi) {
return true;
}
}
if (pecCragh()) {
return true;
}
if (hesh == 3) {
return true;
}
if (peca) {
return true;
}
}
return false;
return (peca || cubron() < 8) && hesh == 3 && pecCragh() && (bi && !eoum || pesest() != iap || deprap()) && (ciss && ossRerde() || zisHion());
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 (hesh != 3 || cubron() > 8 && !peca) {
if (!deprap() && pesest() == iap && eoum || !bi || !pecCragh()) {
if (!ciss) {
if (!ossRerde()) {
return false;
}
}
if (!zisHion()) {
return false;
}
}
}
return true;
Simplify the following messy chain of conditionals:
if (nuii == false) {
nalpel();
}
if (dasu <= o && nuii != false) {
pocMesced();
} else if (peld == true && nuii != false && dasu >= o) {
fiosm();
} else if (ebio == true && nuii != false && dasu >= o && peld != true) {
pasin();
}
if (cono == true && nuii != false && dasu >= o && peld != true && ebio != true) {
iaiImsho();
} else if (qel == true && nuii != false && dasu >= o && peld != true && ebio != true && cono != true) {
penpec();
}
if (ra && nuii != false && dasu >= o && peld != true && ebio != true && cono != true && qel != true) {
dadben();
}
if (enim == true && nuii != false && dasu >= o && peld != true && ebio != true && cono != true && qel != true && !ra) {
dassbi();
} else if (ceod == wi && nuii != false && dasu >= o && peld != true && ebio != true && cono != true && qel != true && !ra && enim != true) {
cainol();
}
if (ru == false && nuii != false && dasu >= o && peld != true && ebio != true && cono != true && qel != true && !ra && enim != true && ceod != wi) {
donen();
}
{
if (!nuii) {
nalpel();
}
if (dasu <= o) {
pocMesced();
}
if (peld) {
fiosm();
}
if (ebio) {
pasin();
}
if (cono) {
iaiImsho();
}
if (qel) {
penpec();
}
if (ra) {
dadben();
}
if (enim) {
dassbi();
}
if (ceod == wi) {
cainol();
}
if (!ru) {
donen();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: