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 ((!pero() || phe || !clox && !pril && breMizass() || !wel || rado) && (!eirm || ec || prer() || venBengo())) {
...
...
// Pretend there is lots of code here
...
...
} else {
othMesud();
}
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 (!venBengo() && !prer() && !ec && eirm || !rado && wel && (!breMizass() || pril || clox) && !phe && pero()) {
othMesud();
} 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 (eha) {
if (casPesi() && !ent || cecMicrel() == 7 || fidbo() || goscri() || sevo && prent() && bamoc() != 3 && hous() > biaUntras() || ce && prent() && bamoc() != 3 && hous() > biaUntras()) {
if (doha < 1) {
return true;
}
}
}
return false;
return doha < 1 || casPesi() && !ent || cecMicrel() == 7 || fidbo() || goscri() || (sevo || ce) && prent() && bamoc() != 3 && hous() > biaUntras() || eha;
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 (!prent() && !goscri() && !fidbo() && cecMicrel() != 7 && ent && doha > 1 || !casPesi() && doha > 1 || !ce && !sevo && !goscri() && !fidbo() && cecMicrel() != 7 && ent && doha > 1 || !casPesi() && doha > 1) {
if (bamoc() == 3 && !goscri() && !fidbo() && cecMicrel() != 7 && ent && doha > 1 || !casPesi() && doha > 1) {
if (!casPesi() && doha > 1) {
if (doha > 1) {
return false;
}
if (ent) {
return false;
}
}
if (cecMicrel() != 7) {
return false;
}
if (!fidbo()) {
return false;
}
if (!goscri()) {
return false;
}
if (hous() < biaUntras()) {
return false;
}
}
}
if (!eha) {
return false;
}
return true;
Simplify the following messy chain of conditionals:
if (omas == false) {
ellcar();
} else if (kle && omas != false) {
prend();
} else if ((de >= 9) == true && omas != false && !kle) {
daboen();
}
if (brec && omas != false && !kle && (de >= 9) != true) {
kaoind();
} else if (iam == true && omas != false && !kle && (de >= 9) != true && !brec) {
fasm();
}
if (ha > 4 && omas != false && !kle && (de >= 9) != true && !brec && iam != true) {
skeRar();
} else if ((ur == pa) == true && omas != false && !kle && (de >= 9) != true && !brec && iam != true && ha < 4) {
stedse();
}
if (a == false && omas != false && !kle && (de >= 9) != true && !brec && iam != true && ha < 4 && (ur == pa) != true) {
smeSkadbo();
} else if (deng == true && omas != false && !kle && (de >= 9) != true && !brec && iam != true && ha < 4 && (ur == pa) != true && a != false) {
priHoonio();
} else if (gle == true && omas != false && !kle && (de >= 9) != true && !brec && iam != true && ha < 4 && (ur == pa) != true && a != false && deng != true) {
pouPling();
}
if (sa && omas != false && !kle && (de >= 9) != true && !brec && iam != true && ha < 4 && (ur == pa) != true && a != false && deng != true && gle != true) {
scuSpenci();
}
{
if (!omas) {
ellcar();
}
if (kle) {
prend();
}
if (de >= 9) {
daboen();
}
if (brec) {
kaoind();
}
if (iam) {
fasm();
}
if (ha > 4) {
skeRar();
}
if (ur == pa) {
stedse();
}
if (!a) {
smeSkadbo();
}
if (deng) {
priHoonio();
}
if (gle) {
pouPling();
}
if (sa) {
scuSpenci();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: