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 ((os || !ceon || esi) && !(!ke && noic() && (stror() || gior() != 1) && !ho && (ut >= ma || in))) {
...
...
// Pretend there is lots of code here
...
...
} else {
cible();
}
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 (!ke && noic() && (stror() || gior() != 1) && !ho && (ut >= ma || in) || !esi && ceon && !os) {
cible();
} 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 (alzun() != sti && !mo && siga() || chea) {
if (shucma() != 8 || e <= taud()) {
if (priet() < 7) {
if (il) {
return true;
}
}
if (iolIpt() == 0) {
return true;
}
}
if (!ent) {
return true;
}
if (acel < armin()) {
return true;
}
}
return false;
return acel < armin() && !ent && (iolIpt() == 0 && (il || priet() < 7) || shucma() != 8 || e <= taud()) || alzun() != sti && !mo && (siga() || chea);
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 (mo && e >= taud() && shucma() == 8 && priet() > 7 && !il || iolIpt() != 0 || ent || acel > armin() || alzun() == sti && e >= taud() && shucma() == 8 && priet() > 7 && !il || iolIpt() != 0 || ent || acel > armin()) {
if (ent || acel > armin()) {
if (iolIpt() != 0) {
if (!il) {
return false;
}
if (priet() > 7) {
return false;
}
}
if (shucma() == 8) {
return false;
}
if (e >= taud()) {
return false;
}
}
if (!siga()) {
return false;
}
if (!chea) {
return false;
}
}
return true;
Simplify the following messy chain of conditionals:
if (i == false) {
caturm();
} else if (ress == true && i != false) {
miobem();
}
if (esm == true && i != false && ress != true) {
sweEltos();
} else if (sas && i != false && ress != true && esm != true) {
meci();
}
if (fel == false && i != false && ress != true && esm != true && !sas) {
eard();
}
if (a && i != false && ress != true && esm != true && !sas && fel != false) {
pril();
}
if (al == 3 && i != false && ress != true && esm != true && !sas && fel != false && !a) {
placi();
} else if (jic < 1 && i != false && ress != true && esm != true && !sas && fel != false && !a && al != 3) {
peau();
} else if (oc == true && i != false && ress != true && esm != true && !sas && fel != false && !a && al != 3 && jic > 1) {
prea();
} else if (op && i != false && ress != true && esm != true && !sas && fel != false && !a && al != 3 && jic > 1 && oc != true) {
bincos();
}
{
if (!i) {
caturm();
}
if (ress) {
miobem();
}
if (esm) {
sweEltos();
}
if (sas) {
meci();
}
if (!fel) {
eard();
}
if (a) {
pril();
}
if (al == 3) {
placi();
}
if (jic < 1) {
peau();
}
if (oc) {
prea();
}
if (op) {
bincos();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: