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 (!ro || gloe() || hibang() && deriwn()) {
...
...
// Pretend there is lots of code here
...
...
} else {
risTelnig();
}
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 ((!deriwn() || !hibang()) && !gloe() && ro) {
risTelnig();
} 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 (!da && aie <= 6 || nicmol() || ol <= pelJacpri() && aie <= 6 || nicmol()) {
if (!su) {
return true;
}
}
return false;
return !su || (!da || ol <= pelJacpri()) && (aie <= 6 || nicmol());
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 (ol >= pelJacpri() && da && su) {
if (su) {
return false;
}
if (aie >= 6) {
return false;
}
if (!nicmol()) {
return false;
}
}
return true;
Simplify the following messy chain of conditionals:
if (rhic == true) {
phun();
}
if (so < 5 && rhic != true) {
ensno();
} else if (stek == 2 && rhic != true && so > 5) {
idon();
} else if (rhic != true && so > 5 && stek != 2) {
vecReic();
}
{
if (rhic) {
phun();
}
if (so < 5) {
ensno();
}
if (stek == 2) {
idon();
}
vecReic();
}
Things to double-check in your solution:
== true and == false checks?else, no final if.Related puzzles: