Re: pointers and assignment operators, basic question
Jeff Bender <jigaboophelps@gmail.com> wrote in
news:1180739875.214500.293340@g4g2000hsf.googlegroups.com:
I am trying to remember how to code in C++ after many years of using
Java exclusively.
I have this setup:
class Base {
public:
virtual void printA(){}
};
class D1 : public Base {
public:
D1() {
a = 1;
}
int a;
void printA() {cout << a << endl;}
};
class D2 : public Base {
public:
D2() {
a = 2;
}
int a;
void printA() {cout << a << endl;}
};
int main() {
D1 * d1 = new D1;
D2 * d2 = new D2;
Base * b1 = d1;
Base * b2 = d2;
*b1 = *b2; // HERE What does this _do_?
Nothing. It attempts to call Base::operator=() and passing it a Base& to
*b2, which by default does a memberwise assignment of the Base portion.
Since Base has no members, it does nothing.
b1->printA();
return 0;
}
The program, as _you_ would expect, outputs 1. _I_ am trying to
figure out why it doesn't output 2.
What does the line marked HERE do? I expected it to overwrite the
memory that starts at b1 with the memory that starts at b2, but that
is clearly not the case or the output would be 2. If I do something
like this:
Uh, no. It does memberwise assignment, not a memcpy.
int * a = new int(3);
int * b = new int(5);
*a = *b;
the memory starting at a was overwritten by the contents of the memory
in b. Why is it different in the situation above? What am I
missing? Thanks for helping me to remember this stuff.
No, this assigns an int to an int. There's a difference between:
*a = *b;
and
memcpy(a, b, sizeof(*a));
(Granted, not much different for an int, could be _wildly_ different for
a class.)