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 ((ciam > 5 && !siot || rangmu() >= 8) && du == 5 && (!(scru() && feec >= 4 && po) || laa && !cheup())) {
...
...
// Pretend there is lots of code here
...
...
} else {
surm();
}
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 ((cheup() || !laa) && scru() && feec >= 4 && po || du != 5 || rangmu() <= 8 && (siot || ciam < 5)) {
surm();
} 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 (hedse() > lol && ol || snoe() && angshu() != iciss() || opsen()) {
if (eou || ranour() <= 6) {
if (lesm && epri() == sor) {
if (epri() == sor) {
return true;
}
if (!nede) {
return true;
}
}
}
}
return false;
return (!nede || lesm) && epri() == sor || eou || ranour() <= 6 || hedse() > lol && ol || snoe() && angshu() != iciss() || opsen();
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 (!snoe() && !ol && ranour() >= 6 && !eou && epri() != sor || !lesm && nede || hedse() < lol && ranour() >= 6 && !eou && epri() != sor || !lesm && nede) {
if (hedse() < lol && ranour() >= 6 && !eou && epri() != sor || !lesm && nede) {
if (!lesm && nede) {
if (epri() != sor) {
return false;
}
}
if (!eou) {
return false;
}
if (ranour() >= 6) {
return false;
}
if (!ol) {
return false;
}
}
if (angshu() == iciss()) {
return false;
}
}
if (!opsen()) {
return false;
}
return true;
Simplify the following messy chain of conditionals:
if (si != thit) {
iareod();
} else if (orp == false && si == thit) {
gress();
}
if (egi == 7 && si == thit && orp != false) {
mikNarfli();
}
if (co == true && si == thit && orp != false && egi != 7) {
ibless();
} else if ((mak == 8) == true && si == thit && orp != false && egi != 7 && co != true) {
cilcin();
}
if (rer == true && si == thit && orp != false && egi != 7 && co != true && (mak == 8) != true) {
cliwo();
} else if (ar == true && si == thit && orp != false && egi != 7 && co != true && (mak == 8) != true && rer != true) {
iompin();
} else if (hoph == true && si == thit && orp != false && egi != 7 && co != true && (mak == 8) != true && rer != true && ar != true) {
ensass();
}
if (si == thit && orp != false && egi != 7 && co != true && (mak == 8) != true && rer != true && ar != true && hoph != true) {
ipchri();
}
{
if (si != thit) {
iareod();
}
if (!orp) {
gress();
}
if (egi == 7) {
mikNarfli();
}
if (co) {
ibless();
}
if (mak == 8) {
cilcin();
}
if (rer) {
cliwo();
}
if (ar) {
iompin();
}
if (hoph) {
ensass();
}
ipchri();
}
Things to double-check in your solution:
== true and == false checks?else, no final if.Related puzzles: