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 ((!oceEngra() || eitin() && horan()) && !gigeed() || !(althu() && a || xorSiw()) || he && !bi && gimac()) {
...
...
// Pretend there is lots of code here
...
...
} else {
nacPranog();
}
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 ((!gimac() || bi || !he) && (althu() && a || xorSiw()) && (gigeed() || (!horan() || !eitin()) && oceEngra())) {
nacPranog();
} 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 (pra && cedur() && piamp() <= us || fioss() == 0 || ic != 5 || oanPri() == 8 && !aess && nanjod() && piamp() <= us || fioss() == 0 || ic != 5 || tegn && piamp() <= us || fioss() == 0 || ic != 5) {
if (oanPri() == 8 && !aess && nanjod() && piamp() <= us || fioss() == 0 || ic != 5 || tegn && piamp() <= us || fioss() == 0 || ic != 5) {
if (ic != 5) {
if (fioss() == 0) {
if (piamp() <= us) {
return true;
}
}
}
if (cedur()) {
return true;
}
}
if (suin() == 6) {
return true;
}
}
if (emoght() != 0) {
return true;
}
return false;
return emoght() != 0 && (suin() == 6 || pra) && (cedur() || oanPri() == 8 && !aess && nanjod() || tegn) && (piamp() <= us || fioss() == 0 || ic != 5);
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 (!tegn && !nanjod() && !cedur() || aess && !cedur() || oanPri() != 8 && !cedur() || !pra && suin() != 6 || emoght() == 0) {
if (piamp() >= us) {
return false;
}
if (fioss() != 0) {
return false;
}
if (ic == 5) {
return false;
}
}
return true;
Simplify the following messy chain of conditionals:
if (ar == false) {
eling();
}
if (paa == false && ar != false) {
wadio();
}
if (plil == true && ar != false && paa != false) {
phaCae();
}
if (sclo == true && ar != false && paa != false && plil != true) {
asmlup();
} else if (cla == true && ar != false && paa != false && plil != true && sclo != true) {
poen();
} else if (pral && ar != false && paa != false && plil != true && sclo != true && cla != true) {
idto();
}
if (or == true && ar != false && paa != false && plil != true && sclo != true && cla != true && !pral) {
zomEun();
} else if (no == true && ar != false && paa != false && plil != true && sclo != true && cla != true && !pral && or != true) {
liglod();
}
if (khog == true && ar != false && paa != false && plil != true && sclo != true && cla != true && !pral && or != true && no != true) {
ertast();
}
if (onn > 9 && ar != false && paa != false && plil != true && sclo != true && cla != true && !pral && or != true && no != true && khog != true) {
laream();
}
{
if (!ar) {
eling();
}
if (!paa) {
wadio();
}
if (plil) {
phaCae();
}
if (sclo) {
asmlup();
}
if (cla) {
poen();
}
if (pral) {
idto();
}
if (or) {
zomEun();
}
if (no) {
liglod();
}
if (khog) {
ertast();
}
if (onn > 9) {
laream();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: