Re: pointers and assignment operators, basic question
On Jun 2, 3:17 am, Jeff Bender <jigaboophe...@gmail.com> wrote:
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_?
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:
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.
assume:
int *a;
int *b;
remark1:
a=b;//this means that from now on 'a' points to where 'b' used to
point.
/*now a==b*/
remark2:
*a=1;//sets the value of the memmory location pointed to by 'a' to 1
*a=*b;/*sets the value of the memmory location pointed to by 'a' to
the value stored at the memmory location pointed to by 'b' .In general
this means that 'a!=b' but '*a==*b' for a while(unless you modify 'a'
or 'b') */
What does the line marked HERE do? I expected it to overwrite the
It does not care what 'b1' and 'b2' actually point to.It just copies a
'Base' object(not a 'D1' nor a 'D2') from the location pointed to by
'b2' to the location pointed to by 'b1' .
figure out why it doesn't output 2.
since the type of 'b1' is still considered to be 'Base' nothing is
printed to the output.
*b1 = *b2; // HERE What does this _do_?
try this one:
b1 = b2;//now the next line prints 2 to the output.
remark3:
the meaning of '*' in pointer delarations and pointer-cast operators
differs from the meaning of '*' in dereferencing statements.
regards,
FM