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 (nas && tre && (me || crar || xede() || !cori)) {
...
...
// Pretend there is lots of code here
...
...
} else {
speOmeor();
}
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 (cori && !xede() && !crar && !me || !tre || !nas) {
speOmeor();
} 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 (ampe && odur && driRorin() || is || scue()) {
if (scue()) {
if (is) {
if (driRorin()) {
return true;
}
}
}
if (odur) {
return true;
}
if (!faru) {
return true;
}
}
if (!iffe) {
return true;
}
return false;
return !iffe && (!faru || ampe) && odur && (driRorin() || is || scue());
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 (!odur || !ampe && faru || iffe) {
if (!driRorin()) {
return false;
}
if (!is) {
return false;
}
if (!scue()) {
return false;
}
}
return true;
Simplify the following messy chain of conditionals:
if (dai == true) {
bocing();
} else if (inlo == false && dai != true) {
jinvas();
}
if (lal == true && dai != true && inlo != false) {
snante();
} else if (glul == mi && dai != true && inlo != false && lal != true) {
skaChibir();
}
if (ic == true && dai != true && inlo != false && lal != true && glul != mi) {
tabuen();
} else if (truw && dai != true && inlo != false && lal != true && glul != mi && ic != true) {
hepen();
}
{
if (dai) {
bocing();
}
if (!inlo) {
jinvas();
}
if (lal) {
snante();
}
if (glul == mi) {
skaChibir();
}
if (ic) {
tabuen();
}
if (truw) {
hepen();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: