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 (swace() && (!no || (!(hantma() || casCloal() || uirph() || phus()) || elpre() == 6 || kacHarcer() || he) && oe)) {
...
...
// Pretend there is lots of code here
...
...
} else {
aling();
}
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 ((!oe || !he && !kacHarcer() && elpre() != 6 && (hantma() || casCloal() || uirph() || phus())) && no || !swace()) {
aling();
} 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 (ucsesm() && ve != 9 || plen == kess || miqan() || asvas() || phoi || jaste() == 5) {
if (geaShir() && !plei && tunso() && ve != 9 || plen == kess || miqan() || asvas() || phoi || jaste() == 5) {
if (asvas() || phoi || jaste() == 5) {
if (plen == kess || miqan()) {
if (ve != 9) {
return true;
}
}
}
if (tunso()) {
return true;
}
if (sper) {
return true;
}
}
}
return false;
return ((sper || geaShir() && !plei) && tunso() || ucsesm()) && (ve != 9 || plen == kess || miqan() || asvas() || phoi || jaste() == 5);
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 (!ucsesm() && !tunso() || plei && !sper || !geaShir() && !sper) {
if (ve == 9) {
return false;
}
if (plen != kess) {
return false;
}
if (!miqan()) {
return false;
}
if (!asvas()) {
return false;
}
if (!phoi) {
return false;
}
if (jaste() != 5) {
return false;
}
}
return true;
Simplify the following messy chain of conditionals:
if (oc) {
cojia();
} else if (el == false && !oc) {
wurse();
} else if (i == true && !oc && el != false) {
shic();
}
if (thu == true && !oc && el != false && i != true) {
cenu();
}
if (er && !oc && el != false && i != true && thu != true) {
lermo();
} else if (cieu == true && !oc && el != false && i != true && thu != true && !er) {
pized();
}
if (ir == 9 && !oc && el != false && i != true && thu != true && !er && cieu != true) {
grahe();
} else if (eu > 6 && !oc && el != false && i != true && thu != true && !er && cieu != true && ir != 9) {
sceFos();
} else if (o == true && !oc && el != false && i != true && thu != true && !er && cieu != true && ir != 9 && eu < 6) {
odit();
}
if (wo == true && !oc && el != false && i != true && thu != true && !er && cieu != true && ir != 9 && eu < 6 && o != true) {
petru();
}
{
if (oc) {
cojia();
}
if (!el) {
wurse();
}
if (i) {
shic();
}
if (thu) {
cenu();
}
if (er) {
lermo();
}
if (cieu) {
pized();
}
if (ir == 9) {
grahe();
}
if (eu > 6) {
sceFos();
}
if (o) {
odit();
}
if (wo) {
petru();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: