Default special member binary compatibility

VRonin

Hi
if I have a class:
header:

class Foo : public QObject{
Q_OBJECT
public:
Foo(QObject* parent = nullptr);
~Foo();
};

Source

Foo::Foo(QObject* parent = nullptr)
:QObject(parent){}
Foo::~Foo() = default;

Can I change the destructor to

Foo::~Foo(){
bar();
}

without breaking binary compatibility?

sierdzio

Gut feeling says "no binary compatibility break" but I don't trust it too much. I guess easiest would be to bake Foo into a library, do the change and check if some code using that library will still work after swapping the lib.

Konstantin Tokarev

Why on earth would it break? Binary compatibility is about interfaces, not implementation, and interface didn't change

VRonin

@Konstantin-Tokarev said in Default special member binary compatibility:

Why on earth would it break?

If you don't have a destructor at all (i.e. you leave it implicitly default) and you add it later then you binary compatibility is broken. My question is, is the result different if I make it explicitly default?

Konstantin Tokarev

@VRonin said in Default special member binary compatibility:

If you don't have a destructor at all (i.e. you leave it implicitly default) and you add it later then you binary compatibility is broken

That's wrong. Compatibility is broken if you add virtual destructor to the class which didn't have one. However, in your case implicit destructor is already virtual, because Foo inherits QObject

VRonin

@Konstantin-Tokarev said in Default special member binary compatibility:

That's wrong

Ok, I was told the opposite and I have the maximum respect for your and David's opinion so I don't know who to believe anymore :(