I have a class that derives from a C struct. The class does not do anything special, other than initialization in the constructor, deinitialization function during the destructor, and a few other methods that call into C functions. Basically, it's a run-of-the-mill wrapper. Using GCC, it complained that my destructor was not virtual, so I made it that. Now I run into segfaults.
/* C header file */
struct A
{
/* ... */
}
// My C++ code
class B : public A
{
public:
B() { /* ... init ... */ }
virtual ~B() { /* ... deinit ... */ }
void do()
{
someCFunction(static_cast<A *>(this));
}
};
I was always under the assumption that the static_cast would return the correct pointer to the base class, pruning off the virtual table pointer. So this may not be the case, since I get a segfault in the C function.
By removing the virtual keyword, the code works fine, except that I get a gcc warning. What is the best work around for this? Feel free to enlighten me :).
Both the explicit and implicit conversion to A* are safe. There is neither need for an explicit cast, nor is it going to introduce vtables anywhere, or anything like that. The language would be fundamentally unusable if this were not the case.
I was always under the assumption that the static_cast would return
the correct pointer to the base class, pruning off the virtual table
pointer.
Is absolutely correct.
The destructor need be virtual only if delete ptr; is called where ptr has type A*- or the destructor invoked manually. And it would be A's destructor that would have to be virtual, which it isn't.
Whatever the problem is in your code, it has nothing to do with the code shown. You need to expand your sample considerably.
The destructor of base classes should be virtual. Otherwise, there's a chance that you run into undefined behavior. This is just a speculation, as the code is not enough to tell the actual reason.
Try making the destructor of A virtual and see if it crashes.
Note that a class and a struct are the same thing, other than default access level, so the fact that one's a class and the other a struct has nothing to do with it.
EDIT: If A is a C-struct, use composition instead of inheritance - i.e. have an A member inside of B instead of extending it. There's no point of deriving, since polymorphism is out of the question.
That's not how static_cast works. A pointer to an object continues to be the same pointer, just with a different type. In this case, you're converting a pointer to a derived type (B) into a pointer to the base type (A).
My guess is that casting the pointer does not actually change the pointer value, i.e., it's still pointing to the same memory address, even though it's been cast into an A* pointer type. Remember that struct and class are synonyms in C++.
As #Luchian stated, if you're mixing C and C++, it's better to keep the plain old C structs (and their pointers) as plain old C structs, and use type composition instead of inheritance. Otherwise you're mixing different pointer implementations under the covers. There is no guarantee that the internal arrangement of the C struct and the C++ class are the same.
UPDATE
You should surround the C struct declaration with an extern "C" specification, so that the C++ compiler knows that the struct is a pure C struct:
extern "C"
{
struct A
{
...
};
}
Or:
extern "C"
{
#include "c_header.h"
}
Related
I want to, basically, inherit a C struct in C++ (well, literally). I have:
struct foo { // C-side definition
int sz;
/* whatever */
// no virtual destructor, special mechanism
};
class cxx_class {
/* something here */
// no virtual destructor, no need
};
class derived : public foo /*, public cxx_class */ {
/* some other stuff here */
};
derived instances (in form of foo*) will be passed back to the external library which only knows and uses the foo part of derived (of course). But the problem is that (I assume) the library uses only c-style cast, which are equivalent to reinterpret_cast in C++, the foo in derived must be at the beginning of the memory block.
I wonder if this is defined behavior and does standard guarantee this?
In other words:
EDIT: The two questions are not the same as pointed out by answers and this part is not answered.
Sometimes I use static_cast for downcast. But some of the code uses reinterpret_cast Is it guaranteed the two always give the same answer?
derived instances (in form of foo*) will be passed back to the external library which only knows and uses the foo part of derived (of course).
If your functions take a foo*, it doesn't matter. Even if constructing your types in such a way that foo is not placed at the beginning of derived, the conversion from derived* to foo* happens in the C++ code which does know about the type layout. The external function then doesn't need to worry about it.
does standard guarantee this?
No.
Since you have no virtual methods though, why don't you simply make foo the first member of derived?
Then check with static_assert that the derived class is standard_layout, which is a relatively good guarantee.
The standard has no requirements about the layout of class hierarchies when multiple classes in that hierarchy have non-static data members.
Now, if derived and cxx_class are both empty (no non-static data members), then derived shall be a standard layout type in C++11+.
I was studying Virtual Functions and Pointers. Below code made me to think about, why does one need Virtual Function when we can type cast base class pointer the way we want?
class baseclass {
public:
void show() {
cout << "In Base\n";
}
};
class derivedclass1 : public baseclass {
public:
void show() {
cout << "In Derived 1\n";
}
};
class derivedclass2 : public baseclass {
public:
void show() {
cout << "In Derived 2\n";
}
};
int main(void) {
baseclass * bptr[2];
bptr[0] = new derivedclass1;
bptr[1] = new derivedclass2;
((derivedclass1*) bptr)->show();
((derivedclass2*) bptr)->show();
delete bptr[0];
delete bptr[1];
return 0;
}
Gives same result if we use virtual in base class.
In Derived 1
In Derived 2
Am I missing something?
Your example appears to work, because there is no data, and no virtual methods, and no multiple inheritance. Try adding int value; to derivedclass1, const char *cstr; to derivedclass2, initialize these in corresponding constructors, and add printing these to corresponding show() methods.
You will see how show() will print garbage value (if you cast pointer to derivedclass1 when it is not) or crash (if you cast the pointer to derivedclass2 when class in fact is not of that type), or behave otherwise oddly.
C++ class member functions AKA methods are nothing more than functions, which take one hidden extra argument, this pointer, and they assume that it points to an object of right type. So when you have an object of type derivedclass1, but you cast a pointer to it to type derivedclass2, then what happens without virtual methods is this:
method of derivedclass2 gets called, because well, you explicitly said "this is a pointer to derivedclass2".
the method gets pointer to actual object, this. It thinks it points to actual instance of derivedclass2, which would have certain data members at certain offsets.
if the object actually is a derivedclass1, that memory contains something quite different. So if method thinks there is a char pointer, but in fact there isn't, then accessing the data it points to will probably access illegal address and crash.
If you instead use virtual methods, and have pointer to common base class, then when you call a method, compiler generates code to call the right method. It actually inserts code and data (using a table filled with virtual method pointers, usually called vtable, one per class, and pointer to it, one per object/instance) with which it knows to call the right method. So when ever you call a virtual method, it's not a direct call, but instead the object has extra pointer to the vtable of the real class, which tells what method should really be called for that object.
In summary, type casts are in no way an alternative to virtual methods. And, as a side note, every type cast is a place to ask "Why is this cast here? Is there some fundamental problem with this software, if it needs a cast here?". Legitimate use cases for type casts are quite rare indeed, especially with OOP objects. Also, never use C-style type casts with object pointers, use static_cast and dynamic_cast if you really need to cast.
If you use virtual functions, your code calling the function doesn't need to know about the actual class of the object. You'd just call the function blindly and correct function would be executed. This is the basis of polymorphism.
Type-casting is always risky and can cause run-time errors in large programs.
Your code should be open for extension but closed for modifications.
Hope this helps.
You need virtual functions where you don't know the derived type until run-time (e.g. when it depends on user input).
In your example, you have hard-coded casts to derivedclass2 and derivedclass1. Now what would you do here?
void f(baseclass * bptr)
{
// call the right show() function
}
Perhaps your confusion stems from the fact that you've not yet encountered a situation where virtual functions were actually useful. When you always know exactly at compile-time the concrete type you are operating on, then you don't need virtual functions at all.
Two other problems in your example code:
Use of C-style cast instead of C++-style dynamic_cast (of course, you usually don't need to cast anyway when you use virtual functons for the problem they are designed to solve).
Treating arrays polymorphically. See Item 3 in Scott Meyer's More Effective C++ book ("Never treat arrays polymorphically").
Can i safely call virtual functions after using static_cast on polymorphic class in situations like in the following code or is it UB?
#include <iostream>
class Base
{
public:
virtual void foo() { std::cout << "Base::foo() \n"; }
};
class Derived : public Base
{
public:
virtual void foo() { std::cout << "Derived::foo() \n"; }
};
int main()
{
Base* derived = new Derived;
Derived* _1 = static_cast<Derived*>(derived);
_1->foo();
}
Yes, you can. Although I don't see the point of doing that in your specific example. Just calling it as
derived->foo();
without any casts would have produced exactly the same effect. I.e. some sort of static_cast in that case would be performed implicitly by the virtual call mechanism.
Note that your static_cast does not in any way suppress the "virtual" nature of the call.
That actually makes me wonder what your question is really about. Why would you even ask about it? What are you trying to do? In your code sample really representative of what you are trying to do?
If the compiler allows you to static_cast and at run-time the dynamic type of the object is as expected, then yes, you can. The question is why do you want to do that...
Yes, but as others have said, you don't need to cast the pointer to the derived type to call virtual functions.
However, it is usually safer to use dynamic_cast when dealing with inherited classes. Using dynamic_cast will generate the proper errors if the type information is incorrect at runtime.
Derived* d = dynamic_cast<Derived*>(derived); //safer, but still unnecessary in this situation
As written, that's going to work, basically because derived is a Derived*. So, all the cast is doing is telling the compiler what you already know. Then again, even without the static cast, you'll just end up with Derived::foo in your output. So, this is somewhat pointless. Still, you might need to do this in a situation where you're absolutely sure you know the actual instanced type of your variable and you need to access some non-virtual members for some reason. If you're using a badly designed class library, for instance...
But, in general, static downcasts are a bad idea. You might end up trying to downcast a variable that isn't a Derived*, in which case, calling virtual (or non-virtual) functions (or, in fact, using that pointer for almost any non-trivial operation) results in Undefined Behavior.
I had a question about C++ destructor behavior, more out of curiosity than anything else. I have the following classes:
Base.h
class BaseB;
class BaseA
{
public:
virtual int MethodA(BaseB *param1) = 0;
};
class BaseB
{
};
Imp.h
#include "Base.h"
#include <string>
class BImp;
class AImp : public BaseA
{
public:
AImp();
virtual ~AImp();
private:
AImp(const AImp&);
AImp& operator= (const AImp&);
public:
int MethodA(BaseB *param1) { return MethodA(reinterpret_cast<BImp *>(param1)); }
private:
int MethodA(BImp *param1);
};
class BImp : public BaseB
{
public:
BImp(std::string data1, std::string data2) : m_data1(data1), m_data2(data2) { }
~BImp();
std::string m_data1;
std::string m_data2;
private:
BImp();
BImp(const BImp&);
BImp& operator= (const BImp&);
};
Now, the issue is that with this code, everything works flawlessly. However, when I make the destructor for BImp virtual, on the call to AImp::MethodA, the class BImp seems to have its data (m_data1 and m_data2) uninitialized. I've checked and made sure the contained data is correct at construction time, so I was wondering what the reason behind this could be...
Cheers!
Edit: param1 was actually a reference to B in MethodA. Looks like I over-sanitized my real code a bit too much!
Edit2: Re-arranged the code a bit to show the two different files. Tested that this code compiles, a well. Sorry about that!
If you are casting between related types as you do in this case, you should use static_cast or dynamic_cast, rather than reinterpret_cast, because the compiler may adjust the object pointer value while casting it to a more derived type. The result of reinterpret_cast is undefined in this case, because it just takes the pointer value and pretends it's another object without any regard for object layout.
MethodA takes its parameters by value. This means a copy is passed (and the copy has to be destroyed). That's my best guess for why you might have a BImpl being destroyed that you didn't expect to be, but I don't see what the virtual or non-virtual nature of A's destructor could possibly have to do with it.
But this code can't compile - you use class B in declaring the virtual function in A, but B isn't defined until later. And I don't know what's going on with that cast - you can't reinterpret_cast class types. Perhaps if you work up a test case which demonstrates your issue, and post that?
There's a lot of iffy stuff in this code, so I'm amazed that it works or compiles in any case.
Passing parameters by value instead of reference to MethodA
Casting a B to a BImp via reinterpret_cast -- bad idea! If you're going to cast in that direction, dynamic_cast is the safest.
I fail to see how you're supposed to get a BImp out of a B. You are not invoking any constructors, and you have none that could be invoked that would accept a B. Your default constructor for BImp is private, and assigning a B that has no data, casted to a BImp that still has no data, to a BImp, still isn't going to give you any data!
Several comments:
Your base classes should have virtual destructors so the derived class' dtor is called instead of the just the base class dtor when the object is deleted.
MethodA taking a BaseB pointer as a parameter only to have the pointer reinterpreted as a BImp (a derived class of BaseB) is dangerous. There is no guarantee something else other than BImp is passed to MethodA. What would happen if just a BaseB object was to MethodA? Potentially lots of bad things, I would suspect.
I'm guessing your code "works flawlessly" because you only pass BImp to MethodA. If you are only passing BImp to MethodA then make the signature match the intent (this has the added benefit of removing that awful reinterpret call).
Your code is ill-formed. It is not valid C++. In C++ language reinterpret_cast can only be used to cast between pointer types, reference types, to perform pointer-to-integer conversions (in either direction).
In your code you are trying to use reinterpret_cast to convert from type B to type BImp. This is explicitly illegal in C++. If your compiler allows this code, you have to consult your compiler's documentation in order to determine what's going on.
Other replies already mentioned "slicing". Keep in mind that this is nothing more than just a guess about specific non-standard behavior of your specific compiler. It has nothing to do with C++ language.
I am currently debugging a crashlog. The crash occurs because the vtable pointer of a (c++-) object is 0x1, while the rest of the object seems to be ok as far as I can tell from the crashlog.
The program crashes when it tries to call a virtual method.
My question: Under what circumstances can a vtable pointer become null? Does operator delete set the vtable pointer to null?
This occurs on OS X using gcc 4.0.1 (Apple Inc. build 5493).
Could be a memory trample - something writing over that vtable by mistake. There is a nearly infinite amount of ways to "achieve" this in C++. A buffer overflow, for example.
Any kind of undefined behaviour you have may lead to this situation. For example:
Errors in pointer arithmetic or other that make your program write into invalid memory.
Uninitialized variables, invalid casts...
Treating an array polymorphically might cause this as a secondary effect.
Trying to use an object after delete.
See also the questions What’s the worst example of undefined behaviour actually possible? and What are all the common undefined behaviour that a C++ programmer should know about?.
Your best bet is to use a bounds and memory checker, as an aid to heavy debugging.
A very common case: trying to call a pure virtual method from the constructor...
Constructors
struct Interface
{
Interface();
virtual void logInit() const = 0;
};
struct Concrete: Interface()
{
virtual void logInit() const { std::cout << "Concrete" << std::endl; }
};
Now, suppose the following implementation of Interface()
Interface::Interface() {}
Then everything is fine:
Concrete myConcrete;
myConcrete.pure(); // outputs "Concrete"
It's such a pain to call pure after the constructor, it would be better to factorize the code right ?
Interface::Interface() { this->logInit(); } // DON'T DO THAT, REALLY ;)
Then we can do it in one line!!
Concrete myConcrete; // CRASHES VIOLENTLY
Why ?
Because the object is built bottom up. Let's look at it.
Instructions to build a Concrete class (roughly)
Allocate enough memory (of course), and enough memory for the _vtable too (1 function pointer per virtual function, usually in the order they are declared, starting from the leftmost base)
Call Concrete constructor (the code you don't see)
a> Call Interface constructor, which initialize the _vtable with its pointers
b> Call Interface constructor's body (you wrote that)
c> Override the pointers in the _vtable for those methods Concrete override
d> Call Concrete constructor's body (you wrote that)
So what's the problem ? Well, look at b> and c> order ;)
When you call a virtual method from within a constructor, it doesn't do what you're hoping for. It does go to the _vtable to lookup the pointer, but the _vtable is not fully initialized yet. So, for all that matters, the effect of:
D() { this->call(); }
is in fact:
D() { this->D::call(); }
When calling a virtual method from within a Constructor, you don't the full dynamic type of the object being built, you have the static type of the current Constructor invoked.
In my Interface / Concrete example, it means Interface type, and the method is virtual pure, so the _vtable does not hold a real pointer (0x0 or 0x01 for example, if your compiler is friendly enough to setup debug values to help you there).
Destructors
Coincidently, let's examine the Destructor case ;)
struct Interface { ~Interface(); virtual void logClose() const = 0; }
Interface::~Interface() { this->logClose(); }
struct Concrete { ~Concrete(); virtual void logClose() const; char* m_data; }
Concrete::~Concrete() { delete[] m_data; } // It's all about being clean
void Concrete::logClose()
{
std::cout << "Concrete refering to " << m_data << std::endl;
}
So what happens at destruction ? Well the _vtable works nicely, and the real runtime type is invoked... what it means here however is undefined behavior, because who knows what happened to m_data after it's been deleted and before Interface destructor was invoked ? I don't ;)
Conclusion
Never ever call virtual methods from within constructors or destructors.
If it's not that, you're left with a memory corruption, tough luck ;)
My first guess would be that some code is memset()'ing a class object.
This is totaly implementation dependant. However it would be quite safe to assume that after delete some other operation may set the memory space to null.
Other possibilities include overwrite of the memory by some loose pointer -- actually in my case it's almost always this...
That said, you should never try to use an object after delete.