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 (spe != 4 || io || !acho || lepson()) {
...
...
// Pretend there is lots of code here
...
...
} else {
ormhui();
}
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 (!lepson() && acho && !io && spe == 4) {
ormhui();
} 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 (pa || cinPrid() || i == ro) {
if (!zu) {
if (ocer) {
return true;
}
}
}
return false;
return ocer || !zu || pa || cinPrid() || i == ro;
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 (!ocer) {
return false;
}
if (zu) {
return false;
}
if (!pa) {
return false;
}
if (!cinPrid()) {
return false;
}
if (i != ro) {
return false;
}
return true;
Simplify the following messy chain of conditionals:
if (sesh == true) {
pimo();
}
if (los == false && sesh != true) {
ferol();
}
if (inua && sesh != true && los != false) {
intdul();
}
if (poct == false && sesh != true && los != false && !inua) {
scid();
}
{
if (sesh) {
pimo();
}
if (!los) {
ferol();
}
if (inua) {
intdul();
}
if (!poct) {
scid();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: