Re: operator =

"" <>
Thu, 23 Dec 2010 12:28:09 CST
On Dec 23, 1:58 pm, 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?

You should call the assignment operator of base class from derived
class manually. Thus, you need to declare an assignment operator in
MyClass and call the operator=() of MyBase.

Murat Hakan Ardal

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