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 (!(oirHejun() && or >= 1) || (si || ble >= 1) && ec == du && rirvi() < he && (!coga || !kaneet() && !(sanpu() <= 6 || nua))) {
...
...
// Pretend there is lots of code here
...
...
} else {
priass();
}
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 (((sanpu() <= 6 || nua || kaneet()) && coga || rirvi() > he || ec != du || ble <= 1 && !si) && oirHejun() && or >= 1) {
priass();
} 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 (thap() && e && vos && o && !ia && taud() < corock() || gi && losec() > heam && censpe() && taud() < corock() || coiPetch() && taud() < corock()) {
if (gi && losec() > heam && censpe() && taud() < corock() || coiPetch() && taud() < corock()) {
if (taud() < corock()) {
return true;
}
if (!ia) {
return true;
}
}
if (o) {
return true;
}
if (vos) {
return true;
}
if (e) {
return true;
}
if (mogn() <= 5) {
return true;
}
}
return false;
return (mogn() <= 5 || thap()) && e && vos && o && (!ia || gi && losec() > heam && censpe() || coiPetch()) && taud() < corock();
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 (!thap() && mogn() >= 5) {
if (!o || !vos || !e) {
if (!coiPetch() && !censpe() && ia || losec() < heam && ia || !gi && ia) {
if (taud() > corock()) {
return false;
}
}
}
}
return true;
Simplify the following messy chain of conditionals:
if (ha == true) {
enoRalphe();
} else if (mi == true && ha != true) {
beto();
}
if (zo == thel == true && ha != true && mi != true) {
disdir();
} else if (sne > cas && ha != true && mi != true && zo == thel != true) {
stesta();
} else if (bia == 5 && ha != true && mi != true && zo == thel != true && sne < cas) {
natand();
}
if (po == false && ha != true && mi != true && zo == thel != true && sne < cas && bia != 5) {
cral();
}
if (ta == false && ha != true && mi != true && zo == thel != true && sne < cas && bia != 5 && po != false) {
ollgi();
} else if (eged == true && ha != true && mi != true && zo == thel != true && sne < cas && bia != 5 && po != false && ta != false) {
tonche();
} else if (crel == true && ha != true && mi != true && zo == thel != true && sne < cas && bia != 5 && po != false && ta != false && eged != true) {
gliScoc();
}
if (dac == false && ha != true && mi != true && zo == thel != true && sne < cas && bia != 5 && po != false && ta != false && eged != true && crel != true) {
cooss();
}
{
if (ha) {
enoRalphe();
}
if (mi) {
beto();
}
if (zo == thel) {
disdir();
}
if (sne > cas) {
stesta();
}
if (bia == 5) {
natand();
}
if (!po) {
cral();
}
if (!ta) {
ollgi();
}
if (eged) {
tonche();
}
if (crel) {
gliScoc();
}
if (!dac) {
cooss();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: