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 (i || !chiu || dete || ard || ested() != aef || !(iest == charou())) {
...
...
// Pretend there is lots of code here
...
...
} else {
kehon();
}
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 (iest == charou() && ested() == aef && !ard && !dete && chiu && !i) {
kehon();
} 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 (se == me && we && ejar == 0 || !cird || dass) {
if (suel()) {
if (ea) {
return true;
}
}
}
return false;
return ea || suel() || se == me && we && ejar == 0 || !cird || dass;
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 (!we && !suel() && !ea || se != me && !suel() && !ea) {
if (!ea) {
return false;
}
if (!suel()) {
return false;
}
if (ejar != 0) {
return false;
}
}
if (cird) {
return false;
}
if (!dass) {
return false;
}
return true;
Simplify the following messy chain of conditionals:
if (reae == true) {
nisssa();
}
if (ot == true && reae != true) {
cerNipra();
} else if (omi == true && reae != true && ot != true) {
nahen();
} else if (!cu && reae != true && ot != true && omi != true) {
seged();
} else if ((riol == co) == true && reae != true && ot != true && omi != true && cu) {
mushis();
}
if (reae != true && ot != true && omi != true && cu && (riol == co) != true) {
sacbi();
}
{
if (reae) {
nisssa();
}
if (ot) {
cerNipra();
}
if (omi) {
nahen();
}
if (!cu) {
seged();
}
if (riol == co) {
mushis();
}
sacbi();
}
Things to double-check in your solution:
== true and == false checks?else, no final if.Related puzzles: