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 ((risPren() || (a > 8 && me == 0 || !giss) && hu == 4 && chiou()) && (!finsen() || mue && prid <= cing && !se)) {
...
...
// Pretend there is lots of code here
...
...
} else {
stiac();
}
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 || prid >= cing || !mue) && finsen() || (!chiou() || hu != 4 || giss && (me != 0 || a < 8)) && !risPren()) {
stiac();
} 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 (hari() && ia && phal() && rol && ca && !e || va == 5 && diru) {
if (niePhem() && phal() && rol && ca && !e || va == 5 && diru) {
if (cadban() && phal() && rol && ca && !e || va == 5 && diru) {
if (va == 5 && diru) {
if (!e) {
return true;
}
if (ca) {
return true;
}
if (rol) {
return true;
}
if (phal()) {
return true;
}
}
if (!ecpi) {
return true;
}
}
}
}
return false;
return (!ecpi || cadban() || niePhem() || hari() && ia) && (phal() && rol && ca && !e || va == 5 && diru);
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 (!ia && !niePhem() && !cadban() && ecpi || !hari() && !niePhem() && !cadban() && ecpi) {
if (va != 5 && e || !ca || !rol || !phal()) {
if (!ca || !rol || !phal()) {
if (e) {
return false;
}
}
if (!diru) {
return false;
}
}
}
return true;
Simplify the following messy chain of conditionals:
if (ocox) {
gresm();
}
if (pid == true && !ocox) {
avia();
} else if (ta == false && !ocox && pid != true) {
bicSti();
}
if (aep > 0 && !ocox && pid != true && ta != false) {
irkdin();
} else if (o == true && !ocox && pid != true && ta != false && aep < 0) {
ractu();
} else if (cuin == true && !ocox && pid != true && ta != false && aep < 0 && o != true) {
nasSodius();
} else if (pra != 1 && !ocox && pid != true && ta != false && aep < 0 && o != true && cuin != true) {
slocha();
}
if (ma == false && !ocox && pid != true && ta != false && aep < 0 && o != true && cuin != true && pra == 1) {
mecxi();
} else if (ur == true && !ocox && pid != true && ta != false && aep < 0 && o != true && cuin != true && pra == 1 && ma != false) {
fadpac();
} else if (gisi > od && !ocox && pid != true && ta != false && aep < 0 && o != true && cuin != true && pra == 1 && ma != false && ur != true) {
ormpo();
}
{
if (ocox) {
gresm();
}
if (pid) {
avia();
}
if (!ta) {
bicSti();
}
if (aep > 0) {
irkdin();
}
if (o) {
ractu();
}
if (cuin) {
nasSodius();
}
if (pra != 1) {
slocha();
}
if (!ma) {
mecxi();
}
if (ur) {
fadpac();
}
if (gisi > od) {
ormpo();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: