Re: operator =
Lasse wrote:
Hi!
I use VS2008 and get:
error C2679: binary '=' : no operator found which takes a right-hand
operand of type 'int' (or there is no acceptable conversion)
when compiling code below:
class MyBase {
int a;
public:
MyBase& operator = (int b) { a=b; return *this; }
MyBase& operator << (int b) { a=b; return *this; }
};
class MyClass : public MyBase {
public:
// MyClass& operator = (int b) { *((MyBase*)this)=b; return *this;
}
};
void func()
{
MyBase X;
X = 1;
X << 1;
MyClass Y;
Y = 1; // Error
Y << 1;
}
This is just a stripped example, but my real "problem" is the same.
Why does "operator <<" work in both cases but not "operator =" ?
If I uncomment the line in MyClass everything runs fine.
My idea was to have a base class with a bunch of nice functions and
operators in ONE place
without having to declare the same thing in the inherited class.
What am I missing?
This is because there is always a copy assignment operator present,
that hides inherited assignement operators. Even if you do not
implement a copy assignement operator, your compiler will implicitedly
create one for you. In your case it's
MyClass & MyClass::operator=(const MyClass&);
To use your base class assignement operator, you can make it visible
via a using statement like this:
class MyClass : public MyBase {
public:
// MyClass& operator = (int b) { *((MyBase*)this)=b; return *this; }
using MyBase::operator=;
};
--
Regards,
Sven Eden
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