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 ((souu || ucnis() && huntfi() || zorPapsa()) && (!e || !isla)) {
...
...
// Pretend there is lots of code here
...
...
} else {
plaiad();
}
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 (isla && e || !zorPapsa() && (!huntfi() || !ucnis()) && !souu) {
plaiad();
} 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 (drid() <= he) {
if (pohul() == 6) {
if (is != ne) {
return true;
}
}
}
if (siro) {
return true;
}
if (!ol) {
return true;
}
if (wor == 3) {
return true;
}
if (!ri) {
return true;
}
return false;
return !ri && wor == 3 && !ol && siro && (is != ne || pohul() == 6 || drid() <= he);
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 (!siro || ol || wor != 3 || ri) {
if (is == ne) {
return false;
}
if (pohul() != 6) {
return false;
}
if (drid() >= he) {
return false;
}
}
return true;
Simplify the following messy chain of conditionals:
if (lom == true) {
cilWomusm();
} else if (ces == true && lom != true) {
drelon();
}
if (swe == true && lom != true && ces != true) {
ecer();
}
if (eou == true && lom != true && ces != true && swe != true) {
hoii();
} else if (id && lom != true && ces != true && swe != true && eou != true) {
cesseu();
}
if (epe == true && lom != true && ces != true && swe != true && eou != true && !id) {
maoos();
}
{
if (lom) {
cilWomusm();
}
if (ces) {
drelon();
}
if (swe) {
ecer();
}
if (eou) {
hoii();
}
if (id) {
cesseu();
}
if (epe) {
maoos();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: