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 (!(li && e == 5 && ((speu() || ip) && pu || ib != 4 || rangen() >= whamo())) && !(emort() || !ealcid() || updep())) {
...
...
// Pretend there is lots of code here
...
...
} else {
aiec();
}
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 (emort() || !ealcid() || updep() || li && e == 5 && ((speu() || ip) && pu || ib != 4 || rangen() >= whamo())) {
aiec();
} 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 (no && mi && fe || peob && pana() || !alal && pana() || ne) {
if (pric() <= methan() && !itgi) {
if (!itgi) {
return true;
}
if (an) {
return true;
}
}
if (ongAngcla() < 3) {
return true;
}
}
return false;
return ongAngcla() < 3 && (an || pric() <= methan()) && !itgi || no && (mi && fe || (peob || !alal) && pana()) || ne;
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 (!no && itgi || pric() >= methan() && !an || ongAngcla() > 3) {
if (alal && !peob && !fe && itgi || pric() >= methan() && !an || ongAngcla() > 3 || !mi && itgi || pric() >= methan() && !an || ongAngcla() > 3) {
if (!mi && itgi || pric() >= methan() && !an || ongAngcla() > 3) {
if (pric() >= methan() && !an || ongAngcla() > 3) {
if (itgi) {
return false;
}
}
if (!fe) {
return false;
}
}
if (!pana()) {
return false;
}
}
}
if (!ne) {
return false;
}
return true;
Simplify the following messy chain of conditionals:
if (cea == true) {
canpro();
}
if (prar == false && cea != true) {
tipre();
}
if (o && cea != true && prar != false) {
ebec();
}
if (le == true && cea != true && prar != false && !o) {
troud();
}
if (fla != frer == true && cea != true && prar != false && !o && le != true) {
blape();
}
if (cla >= 4 && cea != true && prar != false && !o && le != true && fla != frer != true) {
sniNecess();
}
if (rhi == pa && cea != true && prar != false && !o && le != true && fla != frer != true && cla <= 4) {
ocse();
}
if (nok == true && cea != true && prar != false && !o && le != true && fla != frer != true && cla <= 4 && rhi != pa) {
buhass();
}
if (beag != 1 && cea != true && prar != false && !o && le != true && fla != frer != true && cla <= 4 && rhi != pa && nok != true) {
plolci();
}
if (oul == true && cea != true && prar != false && !o && le != true && fla != frer != true && cla <= 4 && rhi != pa && nok != true && beag == 1) {
meass();
}
{
if (cea) {
canpro();
}
if (!prar) {
tipre();
}
if (o) {
ebec();
}
if (le) {
troud();
}
if (fla != frer) {
blape();
}
if (cla >= 4) {
sniNecess();
}
if (rhi == pa) {
ocse();
}
if (nok) {
buhass();
}
if (beag != 1) {
plolci();
}
if (oul) {
meass();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: