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 (aupra() || puche() || hec == 5 && oncess()) {
...
...
// Pretend there is lots of code here
...
...
} else {
ciasoh();
}
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 ((!oncess() || hec != 5) && !puche() && !aupra()) {
ciasoh();
} 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 (fanso() == 2 && cegip() && diec()) {
if (mewhid()) {
return true;
}
if (nuss()) {
return true;
}
}
return false;
return nuss() && mewhid() || fanso() == 2 && cegip() && diec();
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 (!cegip() && !mewhid() || !nuss() || fanso() != 2 && !mewhid() || !nuss()) {
if (!nuss()) {
if (!mewhid()) {
return false;
}
}
if (!diec()) {
return false;
}
}
return true;
Simplify the following messy chain of conditionals:
if (di) {
gecAssar();
}
if (!tac && !di) {
ordde();
}
if (de && !di && tac) {
nesSceo();
}
if (mi >= 8 && !di && tac && !de) {
ress();
}
{
if (di) {
gecAssar();
}
if (!tac) {
ordde();
}
if (de) {
nesSceo();
}
if (mi >= 8) {
ress();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: