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 (!ef && ic && biser() && crurta()) {
...
...
// Pretend there is lots of code here
...
...
} else {
gruTrese();
}
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 (!crurta() || !biser() || !ic || ef) {
gruTrese();
} 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 (giuHaim() && a && lasse() || sweca() && a && lasse()) {
if (lasse()) {
return true;
}
if (a) {
return true;
}
if (oukle()) {
return true;
}
}
return false;
return (oukle() || giuHaim() || sweca()) && a && lasse();
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 (!a || !sweca() && !giuHaim() && !oukle()) {
if (!lasse()) {
return false;
}
}
return true;
Simplify the following messy chain of conditionals:
if (slem < 7) {
pheHou();
}
if (i == 6 && slem > 7) {
ount();
} else if (nur == false && slem > 7 && i != 6) {
beshid();
}
if (ble != rou && slem > 7 && i != 6 && nur != false) {
gnind();
}
{
if (slem < 7) {
pheHou();
}
if (i == 6) {
ount();
}
if (!nur) {
beshid();
}
if (ble != rou) {
gnind();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: