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 (!((be == tru || psoo() <= cedEngrum()) && !er) || ja == 9 && dac || geti || unvem() && qin >= clic && pel || ioss()) {
...
...
// Pretend there is lots of code here
...
...
} else {
eturm();
}
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 (!ioss() && (!pel || qin <= clic || !unvem()) && !geti && (!dac || ja != 9) && (be == tru || psoo() <= cedEngrum()) && !er) {
eturm();
} 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 (chu && eout || e) {
if (sceLawa() && is <= oding() && vemi == 6 && !el && qirel() && ni != 5) {
if (ni != 5) {
return true;
}
if (qirel()) {
return true;
}
if (!el) {
return true;
}
if (vemi == 6) {
return true;
}
if (sii) {
return true;
}
if (!ic) {
return true;
}
}
}
return false;
return (!ic && sii || sceLawa() && is <= oding()) && vemi == 6 && !el && qirel() && ni != 5 || chu && eout || e;
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 (!chu && ni == 5 || !qirel() || el || vemi != 6 || is >= oding() && !sii || ic || !sceLawa() && !sii || ic) {
if (vemi != 6 || is >= oding() && !sii || ic || !sceLawa() && !sii || ic) {
if (!qirel() || el) {
if (ni == 5) {
return false;
}
}
}
if (!eout) {
return false;
}
}
if (!e) {
return false;
}
return true;
Simplify the following messy chain of conditionals:
if (ciod == false) {
macdad();
} else if (ke == true && ciod != false) {
napcas();
} else if (assa == true && ciod != false && ke != true) {
wuddec();
} else if (iga == 2 && ciod != false && ke != true && assa != true) {
pafEnt();
} else if (cef == true && ciod != false && ke != true && assa != true && iga != 2) {
uosStec();
} else if (se && ciod != false && ke != true && assa != true && iga != 2 && cef != true) {
bunvos();
} else if (bair == false && ciod != false && ke != true && assa != true && iga != 2 && cef != true && !se) {
epia();
}
if (!ec && ciod != false && ke != true && assa != true && iga != 2 && cef != true && !se && bair != false) {
dacel();
}
if (stis == 0 && ciod != false && ke != true && assa != true && iga != 2 && cef != true && !se && bair != false && ec) {
bliwi();
} else if (ciod != false && ke != true && assa != true && iga != 2 && cef != true && !se && bair != false && ec && stis != 0) {
rhiud();
}
{
if (!ciod) {
macdad();
}
if (ke) {
napcas();
}
if (assa) {
wuddec();
}
if (iga == 2) {
pafEnt();
}
if (cef) {
uosStec();
}
if (se) {
bunvos();
}
if (!bair) {
epia();
}
if (!ec) {
dacel();
}
if (stis == 0) {
bliwi();
}
rhiud();
}
Things to double-check in your solution:
== true and == false checks?else, no final if.Related puzzles: