How to refer to my base class object from derived (without cast?)

[Charlie R]

From the derived class, I want to access my base class - the actual
*object* of the base-class part of this.

One thing that DOES work is to static_cast the (this) pointer into the
base class type, but that seems very sketchy (is this even always
correct?).

Example - how to fix the 2 "fail" lines?

#include <iostream>

class base {
public:
operator int() const { return 123; }
};

class child : public base {
public:
void show() {
std::cout << this->base::operator int() <<
"\n"; // works
std::cout << this->base << "\n"; // fails - I
want to refer to my parent object
base & parentobj = this->base; // fails too
base & parentobj2 = *
static_cast<base*>(this); // works... ugly?
}
};

int main() {
child obj;
obj.show();
}

[Ralf Fassel]

* Charlie R <charl...@wp.pl>

| From the derived class, I want to access my base class - the actual
| *object* of the base-class part of this.
>
| One thing that DOES work is to static_cast the (this) pointer into the
| base class type, but that seems very sketchy (is this even always
| correct?).
>
| Example - how to fix the 2 "fail" lines?

--<snip-snip>--

| std::cout << this->base << "\n"; // fails - I want to refer to my parent object

std::cout << this->operator int() << "\n";
std::cout << operator int() << "\n";

| base & parentobj = this->base; // fails too

base & parentobj = *this;
std::cout << parentobj.operator int() << "\n";

HTH
R'

[Charlie R]

Wed, 29 Mar 2023 12:13:14 +0200, Ralf Fassel:

> * Charlie R <charl...@wp.pl>
> | From the derived class, I want to access my base class - the actual |
> *object* of the base-class part of this.
>>
> | One thing that DOES work is to static_cast the (this) pointer into the
> | base class type, but that seems very sketchy (is this even always |
> correct?).
>>
> | Example - how to fix the 2 "fail" lines?
> --<snip-snip>--
> | std::cout << this->base << "\n"; // fails - I want to refer to my
> parent object
>
> std::cout << this->operator int() << "\n";
> std::cout << operator int()

thanks.
Though, what when we have 2 base classes and both have int() conversion
operator?

Is there no direct way of saying "take my base object X" in an expression?
Is using the static_cast<type_of_base_class*>(this) the valid way?

[Sam]

Charlie R writes:

> > std::cout << this->operator int() << "\n";
> > std::cout << operator int()
>
> thanks.
> Though, what when we have 2 base classes and both have int() conversion
> operator?

base1 &obj = *this;

or

base2 &obj = *this;

Then obj.operator int().

> Is there no direct way of saying "take my base object X" in an expression?
> Is using the static_cast<type_of_base_class*>(this) the valid way?

A cast is the only way to convert X to Y, in an expression.

[Alf P. Steinbach]

On 2023-03-29 12:21 PM, Charlie R wrote:
> Wed, 29 Mar 2023 12:13:14 +0200, Ralf Fassel:
>
>> * Charlie R <charl...@wp.pl>
>> | From the derived class, I want to access my base class - the actual |
>> *object* of the base-class part of this.
>>>
>> | One thing that DOES work is to static_cast the (this) pointer into the
>> | base class type, but that seems very sketchy (is this even always |
>> correct?).
>>>
>> | Example - how to fix the 2 "fail" lines?
>> --<snip-snip>--
>> | std::cout << this->base << "\n"; // fails - I want to refer to my
>> parent object
>>
>> std::cout << this->operator int() << "\n";
>> std::cout << operator int()
>
> thanks.
> Though, what when we have 2 base classes and both have int() conversion
> operator?
>
> Is there no direct way of saying "take my base object X" in an expression?
> Is using the static_cast<type_of_base_class*>(this) the valid way?

You can and should then do

Base::operator int()

... or in external code

obj.Base::operator int()

- Alf