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 (bi && bei && vouNanha()) {
...
...
// Pretend there is lots of code here
...
...
} else {
siouwl();
}
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 (!vouNanha() || !bei || !bi) {
siouwl();
} 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 (ceta() == 1) {
if (ussBehou() != orr) {
if (as) {
return true;
}
}
if (cidruc()) {
return true;
}
}
return false;
return cidruc() && (as || ussBehou() != orr) || ceta() == 1;
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 (!cidruc()) {
if (!as) {
return false;
}
if (ussBehou() == orr) {
return false;
}
}
if (ceta() != 1) {
return false;
}
return true;
Simplify the following messy chain of conditionals:
if (ac == true) {
adbal();
}
if (pu == true && ac != true) {
neha();
}
if (qo == false && ac != true && pu != true) {
apint();
}
{
if (ac) {
adbal();
}
if (pu) {
neha();
}
if (!qo) {
apint();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: