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 (!(aneng() == cepflo()) && tiss || roal && ent && ri && sisco()) {
...
...
// Pretend there is lots of code here
...
...
} else {
paciss();
}
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 ((!sisco() || !ri || !ent || !roal) && (!tiss || aneng() == cepflo())) {
paciss();
} 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 (!ha && sisog() <= cism() && weupid() || ro && sisog() <= cism() && weupid() || oorIpi() && sisog() <= cism() && weupid()) {
if (ma) {
return true;
}
if (zalo == 2) {
return true;
}
}
return false;
return zalo == 2 && ma || (!ha || ro || oorIpi()) && sisog() <= cism() && weupid();
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 (!oorIpi() && !ro && ha && !ma || zalo != 2) {
if (sisog() >= cism() && !ma || zalo != 2) {
if (zalo != 2) {
if (!ma) {
return false;
}
}
if (!weupid()) {
return false;
}
}
}
return true;
Simplify the following messy chain of conditionals:
if (noc) {
sueme();
} else if (dra == true && !noc) {
femin();
}
if (cer > 8 && !noc && dra != true) {
oilfe();
} else if (el <= 4 && !noc && dra != true && cer < 8) {
desshi();
}
if (truo && !noc && dra != true && cer < 8 && el >= 4) {
cioJarsa();
}
if (zapi == false && !noc && dra != true && cer < 8 && el >= 4 && !truo) {
uace();
}
{
if (noc) {
sueme();
}
if (dra) {
femin();
}
if (cer > 8) {
oilfe();
}
if (el <= 4) {
desshi();
}
if (truo) {
cioJarsa();
}
if (!zapi) {
uace();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: