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Why isn't the copy assingnment operator called when the copy construcotr is not available in C

Time:11-09

Why when let's say I have an object declared like this: Obj o1; which is initialized by the default constructor (not very important here, how was o1 initialized, the point is it was initialized) and I create another object in this manner: Obj o2 = o1; the copy constructor is implicitly called, but if I delete the copy constructor, then, I get a compile error. Why the object o1 is not assigned/copied to o2, like here: Obj o1; Obj o2; o2 = o1;? Why the compiler tries to call a constructor in any instance? Is the = operator within Obj o2 = o1; overloaded?

CodePudding user response:

As mentioned in comments, this

Obj o2 = o1;

has nothing to do with assignment. It is a little unfortunate, often confusing, use of = for initialization when otherwise = means assignment.

Also mentioned in comments, the operator= has to assume that the left operator already exists. Consider this somewhat contrived example:

#include <vector>

struct my_vect {
    my_vect() : data(2) {}
    my_vect(const my_vect& other) : data(other.data) {}
    my_vect& operator=(my_vect& other) {
        // this has already been constructed,
        // hence data.size() is 2 already
        data[0] = other.data[0];
        data[1] = other.data[1];
        return *this;
    }
private:
    std::vector<int> data;
};

It is a structure that contains a std::vector whose size is always 2. The vector is initialized upon constructing a my_vect. When assigning one my_vect to another, then data needs not be initialized. Only the values must be copied. Because operator= assumes that the left hand side operator is already properly constructed (it merely copies to data[0] and data[1]) it cannot possibly be used to construct an object (in the example, accessing data[0] or data[1] would be out of bounds).

TL;DR: Constructors construct objects. Assignment assigns to an already existing object. Thats two fundamentally different things.

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