By default, are objects passed by value or by reference?
c++, pass-by-reference, pass-by-value
Solution
`poly`'s values will be copied into `_poly` -- but you will have made an extra copy in the process. A better way to do it is to pass by const reference:
void polynomial::Set(const vector<int>& poly) {
_poly = poly;
}
EDIT I mentioned in comments about copy-and-swap. Another way to implement what you want is
void polynomial::Set(vector<int> poly) {
_poly.swap(poly);
}
This gives you the additional benefit of having the strong exception guarantee instead of the basic guarantee. In some cases the code might be faster, too, but I see this as more of a bonus. The only thing is that this code might be called "harder to read", since one has to realize that there's an implicit copy.
Problem
Coming from C#, where class instances are passed by reference (that is, a copy of the reference is passed when you call a function, instead of a copy of the value), I'd like to know how this works in C++. In the following case, `_poly = poly`, is it copying the value of `poly` to `_poly`, or what? ``` #include <vector> class polynomial { std::vector<int> _poly; public: void Set(std::vector<int> poly) { poly_ = poly; } }; ```