When i have inheritance, does the compiler-generated functions that i usually get (constructor, destructor, assignment operator and copy constructor) are still generated for my classes?
Let's say i have this inheritance: A base class, B which inherits A (public) and C which public inherits B.
My A class has no memory allocation or anything that requires a destructor to be implemented by me, and i'm not implementing a destructor there, when i compile my program will it still create an empty A::~A(){} ?
Same for B and C.. Thank you!
The rule of 5 still applies to each of the classes, independent of their inheritence.
In other words, if B is derived from A, just because A defined their copy constructor, that doesn't affect the generation of Bs copy constructor.
You should, however, be mindful to define a virtual destructor for the base class if needed.
Yes, of course. And the constructor/destructor chained calls are still present (i.e., C destructor will call B destructor which calls A destructor, same in the reverse order for constructors).
Yes, Compiler always generates (constructor, destructor, assignment operator and copy constructor) for the classes , where user have not defined these functions explicitly.
Compiler inserts constructor, destructor,copy constructor and overloaded assignment operator in a class if it is not defined by the user.
But the most important thing is if user defines a parameterized constructor in a class,then compiler will not generate the default constructor and and object creation without any parameter will throw a linker error.
Say for example you have a class A
class A
{
int a;
public:
//.....
//some line of code
//.....
}
If you don't provide any constructor, the compiler will generate a default constructor which does not takes any parameter A(){}.
But if by any chance you declare a parametrized constructor like
A(int i)
{
a = i;
}
The compiler doesn't generate any default constructor and your object creation with no parameter will fail.
A a; ---> This will fail.
A b(10) ---> This will pass.
So thumb of rule is, if your are providing a constructor of your own, always provide the default constructor along with it.
Related
I was going through a code implementation where the intention was to not let anyone make objects of a particular class. Here is the code snippet:
class CantInstantiate
{
CantInstantiate();
CantInstantiate(const CantInstantiate&);
...
};
Is it really required to make the copy constructor private undefined if the default constructor is already made private undefined (provided there is no other constructor)? What is the benefit of preventing copy of an object when we don't have an original object in the first place? Please explain. Thanks in advance.
If the intent is to prevent making instances of a class then you need to ensure you can't call any constructor. If you don't explicitly prohibit copy construction then you're leaving it up to the compiler to decide whether to include one or not, based on the conditions for causing a deleted implicitly-declared copy constructor which are:
- T has non-static data members that cannot be copied (have deleted, inaccessible, or ambiguous copy constructors);
- T has direct or virtual base class that cannot be copied (has deleted, inaccessible, or ambiguous copy constructors);
- T has direct or virtual base class with a deleted or inaccessible destructor;
- T is a union-like class and has a variant member with non-trivial copy constructor;
- T has a data member of rvalue reference type;
- T has a user-defined move constructor or move assignment operator (this condition only causes the implicitly-declared, not the defaulted, copy constructor to be deleted).
So if we assume that you have a class A that attempts to prevent itself being constructed by only prohibiting default construction and not including anything in the list above, it's possible to exploit the implicit copy constructor to gain an instance of A, or even to inherit from it. For example:
class A
{
A(){};
};
struct B : public A {
B() : A(*a)
{}
A* a;
};
B b;
A a (*static_cast<A*>(nullptr));
Now admittedly the code above could produce unexpected behaviour and any decent compiler would produce warnings if you attempted to do it but it does compile. The practicality of such an approach would depend entirely upon what else was declared within A...
I would argue that if the idea is to stop creation then you need to ensure that all methods of construction are prevented. Therefore, it's a much better approach to explicitly delete the: default constructor, copy constructor and move constructor. This sends a much clearer statement of intent and I believe it should prevent all methods of construction.
class CantInstantiate
{
public:
CantInstantiate() = delete;
CantInstantiate(const CantInstantiate&) = delete;
CantInstantiate(CantInstantiate&&) = delete;
...
};
In C++, if we have this class
class Uncopyable{
public:
Uncopyable(){}
~Uncopyable(){}
private:
Uncopyable(const Uncopyable&);
Uncopyable& operator=(const Uncopyable&);
};
and then we have a derived class
class Dervied: private Uncopyable{
};
My question is: why won't this generate a compile time error when the compiler generates the default copy constructor and assignment operators in the derived class ? Won't the generated code try to access base class private members ?
C++11 12.8/7 states "If the class definition does not explicitly declare a copy constructor, one is declared implicitly." so Dervied has an implicitly declared copy constructor. 12.8/11 says:
An implicitly-declared copy/move constructor is an inline public member of its class. A defaulted copy/move constructor for a class X is defined as deleted (8.4.3) if X has:
a variant member with a non-trivial corresponding constructor and X is a union-like class,
a non-static data member of class type M (or array thereof) that cannot be copied/moved because overload resolution (13.3), as applied to M’s corresponding constructor, results in an ambiguity or a function that is deleted or inaccessible from the defaulted constructor,
a direct or virtual base class B that cannot be copied/moved because overload resolution (13.3), as applied to B’s corresponding constructor, results in an ambiguity or a function that is deleted or inaccessible from the defaulted constructor,
any direct or virtual base class or non-static data member of a type with a destructor that is deleted or inaccessible from the defaulted constructor,
for the copy constructor, a non-static data member of rvalue reference type, or
for the move constructor, a non-static data member or direct or virtual base class with a type that does not have a move constructor and is not trivially copyable.
Specifically, the third bullet applies: Dervied has a direct base class Uncopyable that cannot be copied because overload resolution results in a function that is inaccessible from Dervied::Dervied(const Dervied&). As a result Dervied's implicitly declared copy constructor is declared as deleted, resulting in a compile time error if and only if that copy constructor is called.
why won't this generate a compile time error when the compiler generates the default copy constructor and assignment operators in the derived class ?
Because the compiler generates them only when they are needed by the code being compiled. Write some code using the derived class where the copy constructor and/or assignment operator are involved, and you will see the compile-time error you are looking for.
The private in the inheritance makes them private to Derived, it can still see them, classes that use Derived can't.
The derived class will inherit the private copy constructor but will not need to use it unless you copy an object of derived type, as in this example.
The compiler does not auto-generate constructors/operators unless they are used and no other constructor/operator can be used to do that operation (i.e. a copy operation can be used in some situations where a move operation would suffice). The latter statement results in the following set of rules.
Here are the rules to the auto-generation of certain member functions:
Default constructor (if no other constructor is explicitly declared)
Copy constructor if no move constructor or move assignment operator
is explicitly declared. If a destructor is declared generation of a
copy constructor is deprecated.
Move constructor if no copy
constructor, move assignment operator or destructor is explicitly
declared.
Copy assignment operator if no move constructor or move assignment
operator is explicitly declared. If a destructor is declared
generation of a copy assignment operator is deprecated.
Move assignment operator if no copy constructor, copy assignment operator
or destructor is explicitly declared.
Destructor
The list is taken from this Wikipedia page.
One class cannot call private methods on another class, but it can inherit as much as it is coded too. This code just includes the member functions from Uncopyable in Derived.
Imagine if you wrote a class inheriting from std::vector. You can still erase, insert, push_back and do all those sorts of things. Because these are all public or protected vector member functions, they in turn call implementation specific private functions that do the low level things like manage memory. Your code in this derived class couldn't call those memory management functions directly though. This is used to insure the creators of the vector can change the internal details freely without breaking your use of the class.
If your example is what the code actually looks like, then this it is a common pattern used to make things that cannot be copied. It would make code like the following produce a compiler error:
Derived Foo;
Derived Bar;
Foo = Bar
It would also make the code throw an error on the following:
int GetAnswer(Derived bar)
{ /* Do something with a Derived */ }
Derived Foo;
int answer = GetAnser(Foo);
This example fails because a copy of foo is made and passed as the parameter in the function GetAnswer.
There are many reasons why something might not be copyable. The most common I have experienced is that the object manages some low level resource a single file, an opengl context, an audio output, etc... Imagine if you had a class that managed a log file. If it closed the file in the deconstructor, what happens to the original when a copy is destroyed.
Edit: to pass an Uncopyable class to a function, pass it by reference. The Following function does not make a copy:
int GetAnswer(Derived& bar)
{ /* Do something with a Derived */ }
Derived Foo;
int answer = GetAnser(Foo);
It would also cause a compiler error if all the constructor were private and the class was instantiated. But even if all the member function even constructors were private and the class was never instantiated that would be valid compiling code.
Edit: The reason a class with constructor is that there maybe other way to construct it or it maybe have static member functions, or class functions.
Sometimes factories are used to build object which have no obvious constructor. These might have functions to whatever magic is required to make the umakeable class instance. The most common I have seen is just that there was another constructor that was public, but not in an obvious place. I have also seen factories as friend classes to the unconstructable class so they could call the constructors and I have seen code manually twiddle bits of memory and cast pointers to the memory it to an instance of a class. All of these patterns are used to insure that a class is correctly created beyond just the guarantees C++ supplies.
A more common pattern I have seen is static member functions in classes.
class uncreateable
{
uncreateable() {}
public:
static int GetImportantAnswer();
};
Looking at this it can be seen that not only do I not need to create a instance of the class to call GetImportantAnswer() but I couldn't create an instance if I wanted. I could call this code using the following:
int answer;
answer = uncreateable::GetImportantAnswer();
Edit: Spelling and grammar
Well, actually this program does not compile with g++:
#include <iostream>
using namespace std;
class Uncopyable{
public:
Uncopyable(){}
~Uncopyable(){}
private:
Uncopyable(const Uncopyable&) {cout<<"in parent copy constructor";}
Uncopyable& operator=(const Uncopyable&) { cout << "in parent assignment operator";}
};
class Derived: private Uncopyable{
};
int main() {
Derived a;
Derived b = a;
}
compiler output:
$ g++ 23183322.cpp
23183322.cpp:10:88: warning: control reaches end of non-void function [-Wreturn-type]
Uncopyable& operator=(const Uncopyable&) { cout << "in parent assignment operator";}
^
23183322.cpp:13:7: error: base class 'Uncopyable' has private copy constructor
class Derived: private Uncopyable{
^
23183322.cpp:9:5: note: declared private here
Uncopyable(const Uncopyable&) {cout<<"in parent copy constructor";}
^
23183322.cpp:19:15: note: implicit copy constructor for 'Derived' first required here
Derived b = a;
^
1 warning and 1 error generated.
I have a class A.
A has its own destructor.
I use A to define B as follows.
class A{
protected:
int* array;
public:
A(int size){array = new int[size];}
~A() { delete [] array;}
}
class B{
public:
A x;
}
How should I define the destructor for B?
Thanks in advance.
You don't have to do anything special with class B. The compiler generated destructor will do the right thing in this case.
class A on the other hand is either missing an assignment operator and a copy constructor, or you have to disable these by making them private. The compiler generated ones will cause problems if you copy or assign an A instance (and hence, a B instance too. See the rule of three.
You don't actually need to define a destructor for B. B doesn't have any members which need manual clean up. When an instance of B is destroyed the destructors of all of it's members are called (whether or not you have a user defined destructor), which means the instance of A is cleaned up by it's own destructor. Therefore the default destructor is perfectly fine for B.
Of course you are missing a custom copy constructor and assignment operator for B to satisfy the rule of three, but thats a different matter.
When the compiler generated destructor is called (for which you don't need to do a thing), the destructor for A will automatically be called.
Basically, you don't need to do anything to destruct B properly in this case.
There's nothing B needs to do in its destructor. You do need to either define a default constructor for A or call its existing constructor in Bs default constructor
If the class doesn't have the constructor, will the compiler make one default constructor for it ?
Programmers new to C++ often have two common misunderstandings:
That a default constructor is synthesized for every class that does
not define one
from the book Inside the C++ Object Model
I am at a loss...
This is well explained in the section from which this quote is taken. I will not paraphrase it in its entirety, but here is a short summary of the section content.
First of all, you need to understand the following terms: implicitly-declared, implicitly-defined, trivial, non-trivial and synthesized (a term that is used by Stanley Lippman, but is not used in the standard).
implicitly-declared
A constructor is implicitly-declared for a class if there is no user-declared constructor in this class. For example, this class struct T { }; does not declare any constructor, so the compiler implicitly declares a default constructor. On the other hand, this class struct T { T(int); }; declares a constructor, so the compiler will not declare an implicit default constructor. You will not be able to create an instance of T without parameters, unless you define your own default constructor.
implicitly-defined
An implicitly-declared constructor is implicitly-defined when it is used, i.e. when an instance is created without parameters. Assuming the following class struct T { };, the line T t; will trigger the definition of T::T(). Otherwise, you would have a linker error since the constructor would be declared but not defined. However, an implicitly-defined constructor does not necessarily have any code associated with it! A default constructor is synthesized (meaning that some code is created for it) by the compiler only under certain circumstances.
trivial constructor
An implicitly-declared default constructor is trivial when:
its class has no virtual functions and no virtual base classes and
its base classes have trivial constructors and
all its non-static members have trivial constructors.
In this case, the default compiler has nothing to do, so there is no code synthesized for it. For instance, in the following code
struct Trivial
{
int i;
char * pc;
};
int main()
{
Trivial t;
}
the construction of t does not involve any operations (you can see that by looking at the generated assembly: no constructor is called to construct t).
non-trivial
On the other hand, if the class does not meet the three requirements stated above, its implicitly-declared default constructor will be non-trivial, meaning that it will involve some operations that must be performed in order to respect the language semantics. In this case, the compiler will synthesize an implementation of the constructor performing these operations.
For instance, consider the following class:
struct NonTrivial
{
virtual void foo();
};
Since it has a virtual member function, its default constructor must set the virtual table pointer to the correct value (assuming the implementation use a virtual method table, of course).
Similarly, the constructor of this class
struct NonTrivial
{
std::string s;
};
must call the string default constructor, as it is not trivial. To perform these operations, the compiler generates the code for the default constructor, and calls it anytime you create an instance without parameters. You can check this by looking at the assembly corresponding to this instantiation NonTrivial n; (you should see a function call, unless the constructor has been inlined).
Summary
When you don't provide any constructor for your class, the compiler implicitly declares a default one. If you try to use it, the compiler implicitly defines it, if it can (it is not always possible, for instance when a class has a non-default-constructible member). However, this implicit definition does not imply the generation of any code. The compiler needs to generate code for the constructor (synthesize it) only if it is non-trivial, meaning that it involves certain operations needed to implement the language semantics.
N.B.
Stanley B Lippman's "Inside the C++ object model" and this answer deals with (a possible) implementation of C++, not its semantics. As a consequence, none of the above can be generalized to all compilers: as far as I know, an implementation is perfectly allowed to generate code even for a trivial constructor. From the C++ user point of view, all that matters is the "implicitly-declared/defined` aspect (and also the trivial/non-trivial distinction, as it has some implications (for instance, an object of a class with non-trivial constructor cannot be a member of a union)).
I think the misconception is:
That a default constructor is synthesized for every class that does not define one
That people think the default constructor, which accepts no arguments, will always be generated if you don't declare it yourself.
However, this is not true, because if you declare any constructor yourself, the default one will not be automatically created.
class MyClass {
public:
MyClass(int x) {}; // No default constructor will be generated now
};
This will lead to problems like when beginners expect to use MyClass like this:
MyClass mc;
Which won't work because there is no default constructor that accepts no args.
edit as OP is still a little confused.
Imagine that my MyClass above was this:
class MyClass {
};
int main() {
MyClass m;
}
That would compile, because the compiler will autogenerate the default constructor MyClass() because MyClass was used.
Now take a look at this:
#include <iostream>
class MyClass {
};
int main() {
std::cout << "exiting\n";
}
If this were the only code around, the compiler wouldn't even bother generating the default constructor, because MyClass is never used.
Now this:
#include <iostream>
class MyClass {
public:
MyClass(int x = 5) { _x = x; }
int _x;
};
int main() {
MyClass m;
std::cout << m._x;
}
The compiler doesn't generate default constructor MyClass(), because the class already has a constructor defined by me. This will work, and MyClass(int x = 5) works as your default constructor because it can accept no arguments, but it wasn't generated by the compiler.
And finally, where beginners might run into a problem:
class MyClass() {
public:
MyClass(int x) { _x = x; }
int _x;
};
int main() {
MyClass m;
}
The above will throw you an error during compilation, because MyClass m needs a default constructor (no arguments) to work, but you already declared a constructor that takes an int. The compiler will not generate a no-argument constructor in this situation either.
A default constructor is synthesized for every class that does not define one if:
The code using the class needs one & only if
There is no other constructor explicitly defined for the class by you.
All the upvoted answers thus far seem to say approximately the same thing:
A default constructor is synthesized for every class that does not have any user-defined constructor.
which is a modification of the statement in the question, which means
A default constructor is synthesized for every class that does not have a user-defined default constructor.
The difference is important, but the statement is still wrong.
A correct statement would be:
A default constructor is synthesized for every class that does not have any user-defined constructor and for which all sub-objects are default-constructible in the context of the class.
Here are some clear counter-examples to the first statement:
struct NoDefaultConstructor
{
NoDefaultConstructor(int);
};
class Surprise1
{
NoDefaultConstructor m;
} s1; // fails, no default constructor exists for Surprise1
class Surprise1 has no user-defined constructors, but no default constructor is synthesized.
It doesn't matter whether the subobject is a member or a base:
class Surprise2 : public NoDefaultConstructor
{
} s2; // fails, no default constructor exists for Surprise2
Even if all subobjects are default-constructible, the default constructor has to be accessible from the composite class:
class NonPublicConstructor
{
protected:
NonPublicConstructor();
};
class Surprise3
{
NonPublicConstructor m;
} s3; // fails, no default constructor exists for Surprise3
Yes a default constructor is always there by default if you don't define a constructor of your own (see the default constructor section here).
http://www.codeguru.com/forum/archive/index.php/t-257648.html
Quote:
The following sentense are got from the book "Inside the C++ object model" , written by Stanley B. Lippman.
There are four characteristics of a class under which the compiler
needs to synthesize a default constructor for classes that declare no
constructor at all. The Standard refers to these as implicit,
nontrivial default constructors. The synthesized constructor fulfills
only an implementation need. It does this by invoking member object or
base class default constructors or initializing the virtual function
or virtual base class mechanism for each object. Classes that do not
exhibit these characteristics and that declare no constructor at all
are said to have implicit, trivial default constructors. In practice,
these trivial default constructors are not synthesized. ...
Programmers new to C++ often have two common misunderstandings:
That a default constructor is synthesized for every class that does
not define one
That the compiler-synthesized default constructor provides explicit
default initializers for each data member declared within the class
As you have seen, neither of these is true.
struct Base{
Base(Base &){} // suppress default constructor
};
struct Derived : Base{
};
int main(){
Derived d;
}
The code shown gives error because the default constructor (implicit) of 'Base' is suppressed. Indeed the standard says in $12.1 "If there is no user-declared constructor for class X, a default constructor is implicitly declared."
There are three things:
a) Does the standard say anywhere that
if the user declared constructor is
present in a class, the default
constructor (implicit) is suppressed. It is bascically the above phrased negatively or is it once again implied :)?
b) Why is it that way?
c) Why the same rules do not apply for the default destructor?
I think that a) is sufficiently clearly implied by your quote.
As for “why” – quite simple: a default constructor is not always meaningful; if there were no way to suppress it, this would weaken C++ substantially.
As for c), a class without destructor (no “default”, just plain destructor) is simply not meaningful.
a) Does the standard say anywhere that
if the user declared constructor is
present in a class, the default
constructor (implicit) is suppressed.
It is bascically the above phrased
negatively or is it once again implied
:)?
Yes, that is the meaning
b) Why is it that way?
Most likely, if you have a user-defined constructor, it means special work needs to be done to initialize the object. It makes sense to disable the implicitly generated default constructor in such a case, because it likely won't do any of the special work.
c) Why the same rules do not apply for the default destructor?
Well, perhaps it would make sense for the language to enforce the "rule of three" (if you define one of copy constructor, assignment operator or destructor, chances are you need to implement all three), but it just doesn't.
Perhaps the rationale is that there are several ways to initialize a class, but assignment and destruction often works the same way (memberwise assignment, run destructor of all members).
The shortest answer is because you declared a constructor for the class Base, no default constructor is created (thus the suppression). You cannot initialize Derived because Derived has no default constructor to call on class Base. (this is because the default constructor of derived that is generated for you can only construct it's parent class with the default constructor)