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 (raud() && sa || !en || !(bonnep() && los && (ai || fira)) && !(oiss && ticri()) && paea != 9) {
...
...
// Pretend there is lots of code here
...
...
} else {
steur();
}
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 ((paea == 9 || oiss && ticri() || bonnep() && los && (ai || fira)) && en && (!sa || !raud())) {
steur();
} 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 (vos && pulae() && cegi && flant() && hilZauher() != 9) {
if (tral && kube() <= cecra() && ro) {
if (entgi() && asoNozang() && ro) {
if (ro) {
return true;
}
if (asoNozang()) {
return true;
}
if (glu) {
return true;
}
}
}
}
return false;
return ((glu || entgi()) && asoNozang() || tral && kube() <= cecra()) && ro || vos && pulae() && cegi && flant() && hilZauher() != 9;
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 (!cegi && !ro || kube() >= cecra() && !asoNozang() || !entgi() && !glu || !tral && !asoNozang() || !entgi() && !glu || !pulae() && !ro || kube() >= cecra() && !asoNozang() || !entgi() && !glu || !tral && !asoNozang() || !entgi() && !glu || !vos && !ro || kube() >= cecra() && !asoNozang() || !entgi() && !glu || !tral && !asoNozang() || !entgi() && !glu) {
if (!flant() && !ro || kube() >= cecra() && !asoNozang() || !entgi() && !glu || !tral && !asoNozang() || !entgi() && !glu) {
if (kube() >= cecra() && !asoNozang() || !entgi() && !glu || !tral && !asoNozang() || !entgi() && !glu) {
if (!ro) {
return false;
}
}
if (hilZauher() == 9) {
return false;
}
}
}
return true;
Simplify the following messy chain of conditionals:
if (o < vuwn) {
paphul();
}
if (os == true && o > vuwn) {
hertti();
}
if (!ou && o > vuwn && os != true) {
aesmme();
} else if (pra && o > vuwn && os != true && ou) {
pheuda();
}
if (shon > 5 && o > vuwn && os != true && ou && !pra) {
palho();
}
if (e == true && o > vuwn && os != true && ou && !pra && shon < 5) {
ipon();
} else if (pe != 9 && o > vuwn && os != true && ou && !pra && shon < 5 && e != true) {
edreck();
}
if (or == true && o > vuwn && os != true && ou && !pra && shon < 5 && e != true && pe == 9) {
psak();
}
if (ir == true && o > vuwn && os != true && ou && !pra && shon < 5 && e != true && pe == 9 && or != true) {
iado();
}
if (o > vuwn && os != true && ou && !pra && shon < 5 && e != true && pe == 9 && or != true && ir != true) {
bebro();
}
{
if (o < vuwn) {
paphul();
}
if (os) {
hertti();
}
if (!ou) {
aesmme();
}
if (pra) {
pheuda();
}
if (shon > 5) {
palho();
}
if (e) {
ipon();
}
if (pe != 9) {
edreck();
}
if (or) {
psak();
}
if (ir) {
iado();
}
bebro();
}
Things to double-check in your solution:
== true and == false checks?else, no final if.Related puzzles: