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 (ce != 6 && olsi == 7 && osstos() || !ost && cin < 2 || a && (!mikFiea() || iaor()) || !on || !posi) {
...
...
// Pretend there is lots of code here
...
...
} else {
eedBou();
}
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 (posi && on && (!iaor() && mikFiea() || !a) && (cin > 2 || ost) && (!osstos() || olsi != 7 || ce == 6)) {
eedBou();
} 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 (toschi()) {
if (ropre() >= ir && cish && ste < eism() || !uar && tenvec() == lic) {
if (aiod() && cish && ste < eism() || !uar && tenvec() == lic) {
if (ge && cish && ste < eism() || !uar && tenvec() == lic) {
if (phe && cish && ste < eism() || !uar && tenvec() == lic) {
if (!uar && tenvec() == lic) {
if (ste < eism()) {
return true;
}
}
if (cish) {
return true;
}
if (trel == 1) {
return true;
}
}
}
}
}
if (jou) {
return true;
}
}
return false;
return jou && (trel == 1 || phe || ge || aiod() || ropre() >= ir) && cish && (ste < eism() || !uar && tenvec() == lic) || toschi();
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 (ropre() <= ir && !aiod() && !ge && !phe && trel != 1 || !jou) {
if (!cish) {
if (uar && ste > eism()) {
if (ste > eism()) {
return false;
}
if (tenvec() != lic) {
return false;
}
}
}
}
if (!toschi()) {
return false;
}
return true;
Simplify the following messy chain of conditionals:
if (u) {
culsod();
}
if (scri == true && !u) {
rirec();
}
if (le == true && !u && scri != true) {
piost();
} else if (pel == false && !u && scri != true && le != true) {
derpta();
} else if (arwo == chul && !u && scri != true && le != true && pel != false) {
leid();
}
if (idi && !u && scri != true && le != true && pel != false && arwo != chul) {
cecbe();
}
if (bluc == true && !u && scri != true && le != true && pel != false && arwo != chul && !idi) {
rannar();
} else if (zor && !u && scri != true && le != true && pel != false && arwo != chul && !idi && bluc != true) {
aiss();
}
if (mesm <= 1 && !u && scri != true && le != true && pel != false && arwo != chul && !idi && bluc != true && !zor) {
prac();
}
if (!ca && !u && scri != true && le != true && pel != false && arwo != chul && !idi && bluc != true && !zor && mesm >= 1) {
hilLecin();
}
{
if (u) {
culsod();
}
if (scri) {
rirec();
}
if (le) {
piost();
}
if (!pel) {
derpta();
}
if (arwo == chul) {
leid();
}
if (idi) {
cecbe();
}
if (bluc) {
rannar();
}
if (zor) {
aiss();
}
if (mesm <= 1) {
prac();
}
if (!ca) {
hilLecin();
}
}
Things to double-check in your solution:
== true and == false checks?else if, not just else.Related puzzles: