Some hours ago I read this question in which the user initialized a const object as
const QList warnings = QList() << 0 << 3 << 7;
Indeed, this expression somehow hurt me because going against the concepts I have of const and of overloading operator<<.
I tried to define an operator<< possibly allowing for the mentioned expression and here is what I got:
typedef vector<int> Vect;
Vect operator<<(Vect v, const int i){ //ByVal v, This way it works, but at which cost?
v.push_back(i); //v is copied at every call!
return v;
};
//Here is where I build the object
const Vect const_vect = Vect() << 1 << 2;
//it works fine but I see massive overhead due to copies..@_@
//Just for testing
int main() {
for(auto e = const_vect.begin(); e!=const_vect.end() ; ++e){
cout << *e << " ";
}
return 0;
}
The previous code (which is also here) works fine.
The following, on the other hand, is the expression I would have expected from the definition of operator<<:
Vect& operator<<(Vect& v, const int i){ //ByRef v<-this I would have expected;
v.push_back(i); //however it can't work because receives and
return v; //returns reference to temporary object.
};
My question is: what am I missing? is the operator<< defined in QList differently implemented from mine, in particular, can it receive and return ByRef?
Secondly, is perhaps the aforementioned a standard procedure to initialize complex const objects?
Thirdly (maybe inconvenient), how would such an expression handled by the compiler? what is done at compile-time and what at run-time?
