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 (acur() >= 9 && !(ar && wrolip() || bu || olha() == henle() && !thel || cheak() && pi >= 0 || pirmot() && fola())) {
...
...
// Pretend there is lots of code here
...
...
} else {
cotvel();
}
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 (ar && wrolip() || bu || olha() == henle() && !thel || cheak() && pi >= 0 || pirmot() && fola() || acur() <= 9) {
cotvel();
} 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 (ocon() && timir() && asza && cas || wube() != cic || iro && atre() && cas || wube() != cic || ir && timir() && asza && cas || wube() != cic || iro && atre() && cas || wube() != cic || knu == deng || apir() < om) {
if (vior == ehon()) {
return true;
}
}
return false;
return vior == ehon() || (ocon() || ir) && timir() && (asza || iro && atre()) && (cas || wube() != cic) || knu == deng || apir() < om;
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 (!ir && !ocon() && vior != ehon()) {
if (!timir() && vior != ehon()) {
if (!atre() && !asza && vior != ehon() || !iro && !asza && vior != ehon()) {
if (vior != ehon()) {
return false;
}
if (!cas) {
return false;
}
if (wube() == cic) {
return false;
}
}
}
}
if (knu != deng) {
return false;
}
if (apir() > om) {
return false;
}
return true;
Simplify the following messy chain of conditionals:
if (!am) {
thriop();
}
if (u == false && am) {
ents();
}
if (pasm && am && u != false) {
nengci();
} else if (pe == false && am && u != false && !pasm) {
edgnen();
}
if ((un == teac) == true && am && u != false && !pasm && pe != false) {
sontsa();
} else if (leen == true && am && u != false && !pasm && pe != false && (un == teac) != true) {
ankqan();
}
if (ri == false && am && u != false && !pasm && pe != false && (un == teac) != true && leen != true) {
posi();
}
if (ma <= 7 && am && u != false && !pasm && pe != false && (un == teac) != true && leen != true && ri != false) {
spham();
} else if (tu <= feic && am && u != false && !pasm && pe != false && (un == teac) != true && leen != true && ri != false && ma >= 7) {
chaar();
}
if (wadu && am && u != false && !pasm && pe != false && (un == teac) != true && leen != true && ri != false && ma >= 7 && tu >= feic) {
metam();
}
{
if (!am) {
thriop();
}
if (!u) {
ents();
}
if (pasm) {
nengci();
}
if (!pe) {
edgnen();
}
if (un == teac) {
sontsa();
}
if (leen) {
ankqan();
}
if (!ri) {
posi();
}
if (ma <= 7) {
spham();
}
if (tu <= feic) {
chaar();
}
if (wadu) {
metam();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: