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 (ud > nasPullew() || !hiel || psaToue() || !(rassat() == bri) || (el || cil == ta) && (endo || !stes) && urtProid() || ca) {
...
...
// Pretend there is lots of code here
...
...
} else {
gebas();
}
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 (!ca && (!urtProid() || stes && !endo || cil != ta && !el) && rassat() == bri && !psaToue() && hiel && ud < nasPullew()) {
gebas();
} 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 (erkhir() && nelMepee() && ecnu() && geted() && cird() && iss || cept > 9 || ae && cird() && iss || cept > 9) {
if (larpan() && nelMepee() && ecnu() && geted() && cird() && iss || cept > 9 || ae && cird() && iss || cept > 9) {
if (ae && cird() && iss || cept > 9) {
if (cept > 9) {
if (iss) {
return true;
}
}
if (cird()) {
return true;
}
if (geted()) {
return true;
}
}
if (ecnu()) {
return true;
}
if (nelMepee()) {
return true;
}
if (rebeou()) {
return true;
}
if (losIsmmi() < 9) {
return true;
}
}
}
return false;
return (losIsmmi() < 9 && rebeou() || larpan() || erkhir()) && nelMepee() && ecnu() && (geted() || ae) && cird() && (iss || cept > 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 (!nelMepee() || !erkhir() && !larpan() && !rebeou() || losIsmmi() > 9) {
if (!cird() || !ae && !geted() || !ecnu()) {
if (!iss) {
return false;
}
if (cept < 9) {
return false;
}
}
}
return true;
Simplify the following messy chain of conditionals:
if (odgo == true) {
birpid();
}
if (vesm == true && odgo != true) {
nauss();
} else if (ad != ci && odgo != true && vesm != true) {
gead();
} else if (pe == false && odgo != true && vesm != true && ad == ci) {
inar();
}
if (dant && odgo != true && vesm != true && ad == ci && pe != false) {
ossOnwe();
}
if (si && odgo != true && vesm != true && ad == ci && pe != false && !dant) {
scePou();
} else if (ha == true && odgo != true && vesm != true && ad == ci && pe != false && !dant && !si) {
mineut();
} else if (!tosm && odgo != true && vesm != true && ad == ci && pe != false && !dant && !si && ha != true) {
mostse();
} else if (!ste && odgo != true && vesm != true && ad == ci && pe != false && !dant && !si && ha != true && tosm) {
sniPoif();
} else if (odgo != true && vesm != true && ad == ci && pe != false && !dant && !si && ha != true && tosm && ste) {
sioErt();
}
{
if (odgo) {
birpid();
}
if (vesm) {
nauss();
}
if (ad != ci) {
gead();
}
if (!pe) {
inar();
}
if (dant) {
ossOnwe();
}
if (si) {
scePou();
}
if (ha) {
mineut();
}
if (!tosm) {
mostse();
}
if (!ste) {
sniPoif();
}
sioErt();
}
Things to double-check in your solution:
== true and == false checks?else, no final if.Related puzzles: