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 ((po || o) && steaa()) {
...
...
// Pretend there is lots of code here
...
...
} else {
astcos();
}
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 (!steaa() || !o && !po) {
astcos();
} 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 (hiat() > 8 && el != i) {
if (umving() == bi && el != i) {
if (el != i) {
return true;
}
if (fesi > sejid()) {
return true;
}
}
}
return false;
return (fesi > sejid() || umving() == bi || hiat() > 8) && el != i;
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 (hiat() < 8 && umving() != bi && fesi < sejid()) {
if (el == i) {
return false;
}
}
return true;
Simplify the following messy chain of conditionals:
if (etar == true) {
mammus();
}
if (eas == true && etar != true) {
dedRemed();
} else if (glir == true && etar != true && eas != true) {
dafSteodi();
}
{
if (etar) {
mammus();
}
if (eas) {
dedRemed();
}
if (glir) {
dafSteodi();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: