pointers and assignment operators, basic question

[Jeff Bender]

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.

 

[Andre Kostur]

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:perator=() 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.)

 

[Sarath]

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.

The assignment really invokes the operator= function generated by the
compiler, which does a bitwise copy.

Copy constructor, operator=, default constructor, destructor functions
will be generated by compiler if we do not write our own.

Regards.
Sarath

 

[John Harrison]

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.

I think you are missing that operator= is not virtual. So because you
call it with Base object all you are doing is calling the Base:perator=.

john

 

[terminator]

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

 

[BobR]

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;}
};

You should prefer 'initializer lists' (vs. assignment) in constructors:

class D1 : public Base { public:
D1() : a( 1 ){} // note the colon
int a;
void printA(){ cout << a << endl;}
};

 

[Jeff Bender]

Everyone, thanks a lot for your help. I understand now. I had also
completely forgotten about initializer lists, but I knew something
about my constructor didn't look right.