Re: operator =

Sven Eden <>
Thu, 23 Dec 2010 12:29:01 CST
Lasse wrote:

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;
  MyBase& operator = (int b) { a=b; return *this; }
  MyBase& operator << (int b) { a=b; return *this; }

class MyClass : public MyBase {
// 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 {
// MyClass& operator = (int b) { *((MyBase*)this)=b; return *this; }
 using MyBase::operator=;

Sven Eden

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