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#Weird Syntax Question
old quartzBOT
hasty shale
You don't need a return statement for a void function unless you want to preemptively exit out of the function
mellow frost
Also it does not make sense to have a comma after a. You’d want a semicolon
magic mirage
Yeah the simple answer is just use braces and semicolons
mellow frost
if(a) cout << a, return;
You can't abuse the comma operator for this
Not much else to say
The continue one is definitely not possible
hasty shale
You could do:
T f(int a) {
if (a) { return std::cout << a, <some_value>; }
return <some_other_value>;
}
Return you can "cheese" yes
mellow frost
Just do if(a) { cout << a; }
Much better than the cursed thing you are doing with the comma
hasty shale
And for a void function you can just do:
void f(int a) {
if (a) std::cout << a;
}
No fun allowed 
mellow frost
Sure, but if I saw that pushed in a PR in my code base I’d get me mallet. 
hasty shale
The continue code of yours is just an endless loop
mellow frost
#cursed-code moment
hasty shale
That doesn't work.
The comma operator expects expressions as each sequence point.
contine is not an expression, nor is return
hasty shale
You can restructure your code, smth like:
void f(int a ){
while(a == 1){
std::cout << "A became 1!";
}
a = 1;
}
my god, then just remove them
magic mirage
#define NOT_A_CURLY_BRACE {
full needle
yes, but I want a continue
hasty shale
void f(int a ){
while(a == 1)
std::cout << "A became 1!";
a = 1;
}
dude
the code was dummy
the main thing I want is to be able to use continue
here I'll write a better one
hasty shale
that'd be great
full needle
void f(int a){
int cnt=0;
while(1){
if(cnt!=5) continue;
doSomeImportantWorkOmg();
break;
}
}
but now
say it has to be like
void f(int a){
int cnt=0, f=0;
while(1){
if(cnt!=5) f^=cnt, continue;
doSomeImportantWorkOmg(f);
break;
}
}
anyway I just want no curly braces and a continue 😭 too much to ask?
hasty shale
```cpp
void f(int a){
int cnt=0;
while(1){
if(cnt!=5) continue;
doSo...
void f(int a){
int cnt=0;
while(cnt != 5)
f ^=cnt;
doSomeImportantWorkOmg();
}
ugh
hasty shale
If you want to forcefully use a continue, there's almost always a better way.
Not to mention that the continue doesn't work with the , operator
Because - as I said before - continue is not an expression (but a statement)
full needle
void solve(){
int n; cin>>n;
vector<int> a(n);
int f=-1, odd=-1;
for(int i=0;i<n;i++) cin>>a[i];
for(int i=0;i<n;i++){
if(a[i]==1) f=i, continue;
// DON'T make this an else if and don't add a a[i]!=1
if(a[i]&1) odd=i;
}
}
I wouldn't know how to do that in a way that'd satisfy you, besides maybe this:
void solve(){
int n; cin>>n;
vector<int> a(n);
int f=-1, odd=-1;
for(int i=0;i<n;i++) cin>>a[i];
for(int i=0;i<n;i++){
cin>>a[i];
if (a[i] & 1)
if (a[i] == 1) odd = i;
else f = i;
}
}
```or some dark `goto` magic, but I would really refrain from that.
But let me tell you, curly braces do not look ugly. Actually many people recommend to put them **even if you only have a oneliner**.😔
hasty shale
If you want to impress them, then do some ternary magic:
for(int i=0;i<n;i++){
cin>>a[i];
f = !(a[i] - 1) ? i : f;
odd = (a[i] - 1) && (a[i] & 1) ? i : odd;
}
@full needle Has your question been resolved? If so, type !solved :)
old quartzBOT
thanks to all involved @here
Did you really just try to ping 32,610 people?
full needle
no just the people in this help channel man
thanks @hasty shale @magic mirage and @mellow frost for your help, I will close this now
!solved
old quartzBOT