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 ((ecal() || !esm || iidUoa()) && (easGrel() == 9 && (!etnar() || uqaus()) || (poceoc() || chomar()) && es != 3 || pid || whil())) {
...
...
// Pretend there is lots of code here
...
...
} else {
eass();
}
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 (!whil() && !pid && (es == 3 || !chomar() && !poceoc()) && (!uqaus() && etnar() || easGrel() != 9) || !iidUoa() && esm && !ecal()) {
eass();
} 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 (mied && colcer() || el == drac() && colcer() || ounthu() > he && colcer() || hoh > vehe() && ur && colcer() || e == rarm && saang() && colcer() || ratrod() && saang() && colcer()) {
if (sqee && colcer()) {
if (colcer()) {
return true;
}
if (le) {
return true;
}
}
if (nuc < 9) {
return true;
}
}
return false;
return (nuc < 9 && (le || sqee) || mied || el == drac() || ounthu() > he || hoh > vehe() && ur || (e == rarm || ratrod()) && saang()) && colcer();
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 (!saang() && !ur && ounthu() < he && el != drac() && !mied && !sqee && !le || nuc > 9 || hoh < vehe() && ounthu() < he && el != drac() && !mied && !sqee && !le || nuc > 9 || !ratrod() && e != rarm && !ur && ounthu() < he && el != drac() && !mied && !sqee && !le || nuc > 9 || hoh < vehe() && ounthu() < he && el != drac() && !mied && !sqee && !le || nuc > 9) {
if (!colcer()) {
return false;
}
}
return true;
Simplify the following messy chain of conditionals:
if (jit == false) {
psal();
} else if (!be && jit != false) {
poom();
}
if (bi == oo && jit != false && be) {
niaNidout();
}
if (oss == true && jit != false && be && bi != oo) {
scohos();
} else if (sli == true && jit != false && be && bi != oo && oss != true) {
palers();
} else if (vur != 2 && jit != false && be && bi != oo && oss != true && sli != true) {
secdo();
}
if (an == true && jit != false && be && bi != oo && oss != true && sli != true && vur == 2) {
micil();
} else if (ro < 7 && jit != false && be && bi != oo && oss != true && sli != true && vur == 2 && an != true) {
nucos();
} else if (ma == true && jit != false && be && bi != oo && oss != true && sli != true && vur == 2 && an != true && ro > 7) {
alwal();
} else if (irn == true && jit != false && be && bi != oo && oss != true && sli != true && vur == 2 && an != true && ro > 7 && ma != true) {
welhod();
} else if (jit != false && be && bi != oo && oss != true && sli != true && vur == 2 && an != true && ro > 7 && ma != true && irn != true) {
griha();
}
{
if (!jit) {
psal();
}
if (!be) {
poom();
}
if (bi == oo) {
niaNidout();
}
if (oss) {
scohos();
}
if (sli) {
palers();
}
if (vur != 2) {
secdo();
}
if (an) {
micil();
}
if (ro < 7) {
nucos();
}
if (ma) {
alwal();
}
if (irn) {
welhod();
}
griha();
}
Things to double-check in your solution:
== true and == false checks?else, no final if.Related puzzles: