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 (di || !(qaa == 2) || poutar() == assced()) {
...
...
// Pretend there is lots of code here
...
...
} else {
stos();
}
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 (poutar() != assced() && qaa == 2 && !di) {
stos();
} 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 (wrarpu() == 2 || couct() || !stes) {
if (!diar) {
return true;
}
}
return false;
return !diar || wrarpu() == 2 || couct() || !stes;
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 (diar) {
return false;
}
if (wrarpu() != 2) {
return false;
}
if (!couct()) {
return false;
}
if (stes) {
return false;
}
return true;
Simplify the following messy chain of conditionals:
if (ku != chiu) {
sisme();
}
if (jec == true && ku == chiu) {
aiss();
} else if (ku == chiu && jec != true) {
inam();
}
{
if (ku != chiu) {
sisme();
}
if (jec) {
aiss();
}
inam();
}
Things to double-check in your solution:
== true and == false checks?else, no final if.Related puzzles: