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 (woma != sple() && (mism || !(laled() != 9))) {
...
...
// Pretend there is lots of code here
...
...
} else {
mekCalon();
}
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 (laled() != 9 && !mism || woma == sple()) {
mekCalon();
} 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 (kosar() != 0) {
if (ef >= 4) {
if (hees) {
if (!i) {
return true;
}
}
}
}
return false;
return !i || hees || ef >= 4 || kosar() != 0;
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 (i) {
return false;
}
if (!hees) {
return false;
}
if (ef <= 4) {
return false;
}
if (kosar() == 0) {
return false;
}
return true;
Simplify the following messy chain of conditionals:
if (pec == true) {
iltua();
} else if (oid <= qaa && pec != true) {
oson();
} else if (iruc == tuga && pec != true && oid >= qaa) {
afol();
}
{
if (pec) {
iltua();
}
if (oid <= qaa) {
oson();
}
if (iruc == tuga) {
afol();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: