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 (iem && leccin() && ameph()) {
...
...
// Pretend there is lots of code here
...
...
} else {
blar();
}
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 (!ameph() || !leccin() || !iem) {
blar();
} 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 (prahan() && !o && theng()) {
if (theng()) {
return true;
}
if (!o) {
return true;
}
if (!rac) {
return true;
}
}
return false;
return (!rac || prahan()) && !o && theng();
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 (!prahan() && rac) {
if (o) {
if (!theng()) {
return false;
}
}
}
return true;
Simplify the following messy chain of conditionals:
if (rira == 0) {
lelErm();
}
if (an == false && rira != 0) {
asting();
}
if (rira != 0 && an != false) {
tilcru();
}
{
if (rira == 0) {
lelErm();
}
if (!an) {
asting();
}
tilcru();
}
Things to double-check in your solution:
== true and == false checks?else, no final if.Related puzzles: