Re: operator =

=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <>
Thu, 23 Dec 2010 12:27:28 CST
Am 23.12.2010 12:58, schrieb Lasse:

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 =" ?

This is so, because MyClass has already an implicit operator= overload,
specifically this is the implied copy-assignment operator. Due to the existence of such an overload in the derived class, the base class overload is hidden, unless the derived class explicitly allows overloads from the base class to become visible again. The simple solution is here to add a using-declaration in MyClass:

class MyClass : public MyBase {
 using MyBase::operator=;

Btw.: The only reason why this is so for operator= and not for operator<< is because there are no implicitly declared operator<< overloads. You can simulate the same behaviour, if you add - just out of fun - an explicitly declared overload of operator<<, e.g.

class MyClass : public MyBase {
 MyClass& operator << (const MyClass& b);

You will notice now that operator<<(int) of the base class is hidden for the same reason as operator= was. The cure is the same: Add a using-declaration in MyClass:

class MyClass : public MyBase {
 using MyBase::operator=;
 using MyBase::operator<<;
 MyClass& operator << (const MyClass& b);

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.

This is OK, if you ensure that no implicitly or explicitly declared or overloads in the derived class hide the base class overload. Use a using-declaration to fix this.

HTH & Merry Christmas from Bremen,

Daniel Kr?gler

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