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 (au < ood && wo && (su && cles != 1 || monec() == maiPidang()) || eldcra() >= 5 && (!ranu() || rac == ril && er == 1) && !be) {
...
...
// Pretend there is lots of code here
...
...
} else {
ossReonau();
}
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 ((be || (er != 1 || rac != ril) && ranu() || eldcra() <= 5) && (monec() != maiPidang() && (cles == 1 || !su) || !wo || au > ood)) {
ossReonau();
} 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 (ce && lalTrer() && ia != 7 && niar <= 1 || cipLiek() && e && niar <= 1 || stuled() && tes < anhial() && ia != 7 && niar <= 1 || cipLiek() && e && niar <= 1) {
if (stuled() && tes < anhial() && ia != 7 && niar <= 1 || cipLiek() && e && niar <= 1) {
if (cipLiek() && e && niar <= 1) {
if (niar <= 1) {
return true;
}
if (ia != 7) {
return true;
}
}
if (lalTrer()) {
return true;
}
}
if (frad != 4) {
return true;
}
if (kadIokep()) {
return true;
}
}
if (en >= tastio()) {
return true;
}
return false;
return en >= tastio() && (kadIokep() && frad != 4 || ce) && (lalTrer() || stuled() && tes < anhial()) && (ia != 7 || cipLiek() && e) && niar <= 1;
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 (!ce && frad == 4 || !kadIokep() || en <= tastio()) {
if (tes > anhial() && !lalTrer() || !stuled() && !lalTrer()) {
if (!e && ia == 7 || !cipLiek() && ia == 7) {
if (niar >= 1) {
return false;
}
}
}
}
return true;
Simplify the following messy chain of conditionals:
if (ha == true) {
daruc();
}
if (prar && ha != true) {
mois();
}
if (oul != 7 && ha != true && !prar) {
etren();
} else if (jeng && ha != true && !prar && oul == 7) {
vose();
} else if (pra == de && ha != true && !prar && oul == 7 && !jeng) {
chlana();
}
if (gru == 8 && ha != true && !prar && oul == 7 && !jeng && pra != de) {
ucapt();
}
if (orm != sa && ha != true && !prar && oul == 7 && !jeng && pra != de && gru != 8) {
bespim();
} else if (rard == false && ha != true && !prar && oul == 7 && !jeng && pra != de && gru != 8 && orm == sa) {
chusm();
} else if (pec == true && ha != true && !prar && oul == 7 && !jeng && pra != de && gru != 8 && orm == sa && rard != false) {
nald();
} else if (ha != true && !prar && oul == 7 && !jeng && pra != de && gru != 8 && orm == sa && rard != false && pec != true) {
orpo();
}
{
if (ha) {
daruc();
}
if (prar) {
mois();
}
if (oul != 7) {
etren();
}
if (jeng) {
vose();
}
if (pra == de) {
chlana();
}
if (gru == 8) {
ucapt();
}
if (orm != sa) {
bespim();
}
if (!rard) {
chusm();
}
if (pec) {
nald();
}
orpo();
}
Things to double-check in your solution:
== true and == false checks?else, no final if.Related puzzles: