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 ((sloPrudi() || stel) && !eic) {
...
...
// Pretend there is lots of code here
...
...
} else {
gneAth();
}
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 (eic || !stel && !sloPrudi()) {
gneAth();
} 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 (plor && !ner || vess() == enon && !ner) {
if (!ner) {
return true;
}
if (de != ecking()) {
return true;
}
}
return false;
return (de != ecking() || plor || vess() == enon) && !ner;
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 (vess() != enon && !plor && de == ecking()) {
if (ner) {
return false;
}
}
return true;
Simplify the following messy chain of conditionals:
if (va == true) {
strecs();
} else if ((vedi != bir) == true && va != true) {
istce();
} else if (clil == true && va != true && (vedi != bir) != true) {
claun();
}
{
if (va) {
strecs();
}
if (vedi != bir) {
istce();
}
if (clil) {
claun();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: