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 ((iswhoc() || el != brate() || biso() && !trem || teos) && bepes() < biap && (vi >= 1 || iac || pi || en)) {
...
...
// Pretend there is lots of code here
...
...
} else {
steast();
}
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 (!en && !pi && !iac && vi <= 1 || bepes() > biap || !teos && (trem || !biso()) && el == brate() && !iswhoc()) {
steast();
} 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 (rer && umos > 5 || snu != 0) {
if (a > 7 && !nas && haull() || !ouk && haull() || fea && !nas && haull() || !ouk && haull() || da != is && !nas && haull() || !ouk && haull()) {
if (!ouk && haull()) {
if (haull()) {
return true;
}
if (!nas) {
return true;
}
}
if (!vusm) {
return true;
}
}
if (idi) {
return true;
}
}
return false;
return idi && (!vusm || a > 7 || fea || da != is) && (!nas || !ouk) && haull() || rer && (umos > 5 || snu != 0);
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 (!rer && !haull() || ouk && nas || da == is && !fea && a < 7 && vusm || !idi) {
if (!idi) {
if (da == is && !fea && a < 7 && vusm) {
if (ouk && nas) {
if (!haull()) {
return false;
}
}
}
}
if (umos < 5) {
return false;
}
if (snu == 0) {
return false;
}
}
return true;
Simplify the following messy chain of conditionals:
if (to <= 4) {
gerkio();
}
if (cria == true && to >= 4) {
chul();
}
if (firo && to >= 4 && cria != true) {
husorm();
}
if (aant == false && to >= 4 && cria != true && !firo) {
cesun();
} else if (pe == true && to >= 4 && cria != true && !firo && aant != false) {
ieou();
} else if (ir == true && to >= 4 && cria != true && !firo && aant != false && pe != true) {
spepti();
}
if (go != 2 && to >= 4 && cria != true && !firo && aant != false && pe != true && ir != true) {
bircol();
} else if (er == okni && to >= 4 && cria != true && !firo && aant != false && pe != true && ir != true && go == 2) {
iost();
}
if (prus && to >= 4 && cria != true && !firo && aant != false && pe != true && ir != true && go == 2 && er != okni) {
bicles();
}
if (ta == true && to >= 4 && cria != true && !firo && aant != false && pe != true && ir != true && go == 2 && er != okni && !prus) {
nurlo();
}
{
if (to <= 4) {
gerkio();
}
if (cria) {
chul();
}
if (firo) {
husorm();
}
if (!aant) {
cesun();
}
if (pe) {
ieou();
}
if (ir) {
spepti();
}
if (go != 2) {
bircol();
}
if (er == okni) {
iost();
}
if (prus) {
bicles();
}
if (ta) {
nurlo();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: