How to assign two different type variable to one variable in c++

Question

I wrote a program that uses a normal Serial port and a Bluetooth Serial Port to send logs to UART and Bluetooth respectively

HardwareSerial Serial;
BluetoothSerial SerialBT;

for example when i want to send something with them i write this

Serial.println("Hello World"); //prints Hello world in normal Serial port
SerialBT.println("Hello World"); // prints Hello World in Bluetooth serial

i defined some preprocessors that checks whether Bluetooth or Normal Serial port is active and if one of them is active print in the active one

my problem is here that i want to declare a variable like "Debugger" that i can assign it with "SerialBT" or "Serial" to write something like this

somewere in my init function

#if BLUETOOTHDEBUG
Debugger=&SerialBT;
#else
Debugger=&Serial;
#endif

and in code

Debbugger->println("Hello World")

instead of using if in whole code

Do these have a common base-class? Could they have one? If so it's easy. If not the answer is basically "no". — tadman
– tadman, Commented Jul 4, 2018 at 19:55
You can always design a wrapper class to expose them in a common interface. — orhtej2
– orhtej2, Commented Jul 4, 2018 at 19:58
HardwareSerial and BluetoothSerial both seem derived from Stream, so perhaps Stream &Debugger should work just fine? — orhtej2
– orhtej2, Commented Jul 4, 2018 at 20:02
Create a minimal reproducible example, because that sounds like you messed up something unrelated. If class B and class C both inherit from class A, then A *a = new B() will work; likewise, if you have B b, then A *a = &b will as well. — anon
– anon, Commented Jul 4, 2018 at 20:31

david · Accepted Answer · 2018-07-04 21:20:31Z

1

based on Comments and the fact that HardwareSerial and BluetoothSerial class are both derived from Stream class, I added

Stream *Debugger

and it worked

answered Jul 4, 2018 at 21:20

david

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