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?
Probably that assignment operators are special, and will be created
for you if you don't provide one. When you comment out your operator,
the compiler will provide
MyClass& operator=(const MyClass&)
which doesn't take an 'int' parameter, just like the compiler says.
You can also simplify you operator= to directly call the base class
operator without using casts:
MyClass& operator=(int b)
{
MyBase::operator=(b);
return *this;
}
Bo Persson
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