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 (osud == pid && (!(ce || plip()) || se <= 8)) {
...
...
// Pretend there is lots of code here
...
...
} else {
kelTovi();
}
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 (se >= 8 && (ce || plip()) || osud != pid) {
kelTovi();
} 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 (cathih() && plo != e && bape >= mual || !o && plo != e && bape >= mual) {
if (bape >= mual) {
return true;
}
if (plo != e) {
return true;
}
if (grinwo()) {
return true;
}
}
return false;
return (grinwo() || cathih() || !o) && plo != e && bape >= mual;
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 (o && !cathih() && !grinwo()) {
if (plo == e) {
if (bape <= mual) {
return false;
}
}
}
return true;
Simplify the following messy chain of conditionals:
if (um == false) {
ninGii();
}
if (ceon == true && um != false) {
sucCiner();
}
if (plen == true && um != false && ceon != true) {
nelol();
} else if (wuro == true && um != false && ceon != true && plen != true) {
sarlma();
}
{
if (!um) {
ninGii();
}
if (ceon) {
sucCiner();
}
if (plen) {
nelol();
}
if (wuro) {
sarlma();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: