Hello Group,
I have a class A, which has three other members and each member is of a
different class
type.
If I don't create a destructor then the compiler will give me one and
whenever class A is
destroyed (whether on stack or on heap) the destructor of each
different object will be called.
However what will happen if the class A defines a destructor like
~A() {} // No code.
The question is that in this case will the destructor of individual
members be called or not,
since the destructor code is emply?
The whole code will look like:
class Object1 {
// Code destructor, constructor
}
class Object2 {
// Code destructor, constructor
}
class Object3 {
// Code destructor, constructor
}
class A {
Object1 o1;
Object2 o2;
Object3 o3;
~A() {} // The question is all about
}
Thanks
nagrik 2 3946
arun wrote: Hello Group,
I have a class A, which has three other members and each member is of a different class type.
If I don't create a destructor then the compiler will give me one and whenever class A is destroyed (whether on stack or on heap) the destructor of each different object will be called.
However what will happen if the class A defines a destructor like
~A() {} // No code.
Same thing really.
Often base classes define this destructor:
virtual ~Base() {}
Same thing as default constructor but is virtual and so this class can
be safely derived. The question is that in this case will the destructor of individual members be called or not, since the destructor code is emply?
Yes. Member variables are destroyed when the lifetime of the
containing object terminates - or is "destroyed" .
arun wrote: Hello Group,
I have a class A, which has three other members and each member is of a different class type.
If I don't create a destructor then the compiler will give me one and whenever class A is destroyed (whether on stack or on heap) the destructor of each different object will be called.
However what will happen if the class A defines a destructor like
~A() {} // No code.
The question is that in this case will the destructor of individual members be called or not, since the destructor code is emply?
The whole code will look like:
class Object1 {
// Code destructor, constructor
} class Object2 {
// Code destructor, constructor
}
class Object3 {
// Code destructor, constructor
}
class A {
Object1 o1; Object2 o2; Object3 o3;
~A() {} // The question is all about
}
Thanks
nagrik
Just remember that when any object goes out of scope its destructor will
be called. Hence when a object goes out of scope, destructor calls for
all the sub objects that make up the object itself will occur. So unless
the destructor needs to be virtual, allocated memory needs to be taken
care of, or the destruction needs to be flagged elsewhere, it's not
absolutely necessary to code for it. Some people prefer to always code
it for completeness
JB This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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Hello Group,
I have a class A, which has three other members and each member is of a
different class
type.
If I don't create a destructor then the compiler will give me one and
whenever class A is
destroyed (whether on stack or on heap) the destructor of each
different object will be called.
|
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Hello Group,
The compiler generated destructor will invoke destructor of each member
of the containing
class let us say 'class A'. However, if I write my own destructor for
class A like;
class Object1 {
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