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 (palk() && (edpic() != 4 || acci) && !coce && sqaStri() && !ic && (eial() || !ze) && endod()) {
...
...
// Pretend there is lots of code here
...
...
} else {
idkoun();
}
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 (!endod() || ze && !eial() || ic || !sqaStri() || coce || !acci && edpic() == 4 || !palk()) {
idkoun();
} 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 (xa && cime != 0 && menk == iard && opri > 2 && besont() >= curvam() || dreark() || el != 5 && !idi || rae < uaasm() && !idi) {
if (el != 5 && !idi || rae < uaasm() && !idi) {
if (dreark()) {
if (besont() >= curvam()) {
return true;
}
if (opri > 2) {
return true;
}
}
if (menk == iard) {
return true;
}
}
if (!prer) {
return true;
}
}
return false;
return (!prer || xa && cime != 0) && (menk == iard && (opri > 2 && besont() >= curvam() || dreark()) || (el != 5 || rae < uaasm()) && !idi);
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 (cime == 0 && prer || !xa && prer) {
if (rae > uaasm() && el == 5 && !dreark() && besont() <= curvam() || opri < 2 || menk != iard) {
if (menk != iard) {
if (opri < 2) {
if (besont() <= curvam()) {
return false;
}
}
if (!dreark()) {
return false;
}
}
if (idi) {
return false;
}
}
}
return true;
Simplify the following messy chain of conditionals:
if (on == false) {
gelap();
} else if (odna == true && on != false) {
sincir();
} else if (te == jied && on != false && odna != true) {
angdim();
}
if (in == true && on != false && odna != true && te != jied) {
cufpit();
}
if (o == true && on != false && odna != true && te != jied && in != true) {
mict();
} else if (efa == true && on != false && odna != true && te != jied && in != true && o != true) {
criOspa();
} else if (prai == har && on != false && odna != true && te != jied && in != true && o != true && efa != true) {
otiTregn();
}
if (ein == true && on != false && odna != true && te != jied && in != true && o != true && efa != true && prai != har) {
phion();
} else if (on != false && odna != true && te != jied && in != true && o != true && efa != true && prai != har && ein != true) {
biss();
}
{
if (!on) {
gelap();
}
if (odna) {
sincir();
}
if (te == jied) {
angdim();
}
if (in) {
cufpit();
}
if (o) {
mict();
}
if (efa) {
criOspa();
}
if (prai == har) {
otiTregn();
}
if (ein) {
phion();
}
biss();
}
Things to double-check in your solution:
== true and == false checks?else, no final if.Related puzzles: