Can someone explains how this virtual table for the different class is stored in memory? When we call a function using pointer how do they make a call to function using address location? Can we get these virtual table memory allocation size using a class pointer? I want to see how many memory blocks is used by a virtual table for a class. How can I see it?
class Base
{
public:
FunctionPointer *__vptr;
virtual void function1() {};
virtual void function2() {};
};
class D1: public Base
{
public:
virtual void function1() {};
};
class D2: public Base
{
public:
virtual void function2() {};
};
int main()
{
D1 d1;
Base *dPtr = &d1;
dPtr->function1();
}
Thanks! in advance
The first point to keep in mind is a disclaimer: none of this is actually guaranteed by the standard. The standard says what the code needs to look like and how it should work, but doesn't actually specify exactly how the compiler needs to make that happen.
That said, essentially all C++ compilers work quite similarly in this respect.
So, let's start with non-virtual functions. They come in two classes: static and non-static.
The simpler of the two are static member functions. A static member function is almost like a global function that's a friend of the class, except that it also needs the class`s name as a prefix to the function name.
Non-static member functions are a little more complex. They're still normal functions that are called directly--but they're passed a hidden pointer to the instance of the object on which they were called. Inside the function, you can use the keyword this to refer to that instance data. So, when you call something like a.func(b);, the code that's generated is pretty similar to code you'd get for func(a, b);
Now let's consider virtual functions. Here's where we get into vtables and vtable pointers. We have enough indirection going on that it's probably best to draw some diagrams to see how it's all laid out. Here's pretty much the simplest case: one instance of one class with two virtual functions:
So, the object contains its data and a pointer to the vtable. The vtable contains a pointer to each virtual function defined by that class. It may not be immediately apparent, however, why we need so much indirection. To understand that, let's look at the next (ever so slightly) more complex case: two instances of that class:
Note how each instance of the class has its own data, but they both share the same vtable and the same code--and if we had more instances, they'd still all share the one vtable among all the instances of the same class.
Now, let's consider derivation/inheritance. As an example, let's rename our existing class to "Base", and add a derived class. Since I'm feeling imaginative, I'll name it "Derived". As above, the base class defines two virtual functions. The derived class overrides one (but not the other) of those:
Of course, we can combine the two, having multiple instances of each of the base and/or derived class:
Now let's delve into that in a little more detail. The interesting thing about derivation is that we can pass a pointer/reference to an object of the derived class to a function written to receive a pointer/reference to the base class, and it still works--but if you invoke a virtual function, you get the version for the actual class, not the base class. So, how does that work? How can we treat an instance of the derived class as if it were an instance of the base class, and still have it work? To do it, each derived object has a "base class subobject". For example, lets consider code like this:
struct simple_base {
int a;
};
struct simple_derived : public simple_base {
int b;
};
In this case, when you create an instance of simple_derived, you get an object containing two ints: a and b. The a (base class part) is at the beginning of the object in memory, and the b (derived class part) follows that. So, if you pass the address of the object to a function expecting an instance of the base class, it uses on the part(s) that exist in the base class, which the compiler places at the same offsets in the object as they'd be in an object of the base class, so the function can manipulate them without even knowing that it's dealing with an object of the derived class. Likewise, if you invoke a virtual function all it needs to know is the location of the vtable pointer. As far as it cares, something like Base::func1 basically just means it follows the vtable pointer, then uses a pointer to a function at some specified offset from there (e.g., the fourth function pointer).
At least for now, I'm going to ignore multiple inheritance. It adds quite a bit of complexity to the picture (especially when virtual inheritance gets involved) and you haven't mentioned it at all, so I doubt you really care.
As to accessing any of this, or using in any way other than simply calling virtual functions: you may be able to come up with something for a specific compiler--but don't expect it to be portable at all. Although things like debuggers often need to look at such stuff, the code involved tends to be quite fragile and compiler-specific.
The virtual table is supposed to be shared between instances of a class. More precisely, it lives at the "class" level, rather than the instance level. Each instance has the overhead of actually having a pointer to the virtual table, if in it's hierarchy there are virtual functions and classes.
The table itself is at least the size necessary to hold a pointer for each virtual function. Other than that, it is an implementation detail how it's actually defined. Check here for a SO question with more details about this.
First of all, the following answer contain almost everything you want to know regarding virtual tables:
https://stackoverflow.com/a/16097013/8908931
If you are looking for something a little more specific (with the regular disclaimer that this might change between platforms, compilers, and CPU architectures):
When needed, a virtual table is being created for a class. The class will have only one instance of the virtual table, and each object of the class will have a pointer which will point to the memory location of this virtual table. The virtual table itself can be thought of as a simple array of pointers.
When you assigned the derived pointer to the base pointer, it also contain the pointer to the virtual table. This mean that the base pointer points to the virtual table of the derived class. The compiler will direct this call to an offset into the virtual table, which will contain the actual address of the function from the derived class.
Not really. Usually at the start of an object, there is a pointer to the virtual table itself. But this will not help you too much, as it is just an array of pointers, with no real indication of its size.
Making a very long answer short: For an exact size you can find this information in the executable (or in segments loaded from it to the memory). With enough knowledge of how the virtual table works, you can get a pretty accurate estimation, given you know the code, the compiler, and the target architecture.
For the exact size, you can find this information in either the executable, or in segments in the memory which are being loaded from the executable. An executable is usually an ELF file, this kind of files, contain information which is needed to run a program. A part of this information is symbols for various kinds of language constructs such as variables, functions and virtual tables. For each symbol, it contains the size it takes in memory. So button line, you will need the symbol name of the virtual table, and enough knowledge in ELF in order to extract what you want.
The answer that Jerry Coffin gave is excellent in explaining how virtual function pointers work to achieve runtime polymorphism in C++. However, I believe that it is lacking in answering where in memory the vtable is stored. As others have pointed out this is not dictated by the standard.
However, there is an excellent blog post(s) by Martin Kysel that goes into great detail about where virtual tables are stored. To summarize the blog post(s):
One vtable is created for every class (not instance) with virtual functions. Each instance of this class points to the same vtable in memory
Each vtable is stored in read only memory of the resulting binary file
The disassembly for each function in the vtable is stored in the text section of the resulting ELF binary
Attempting to write over the vtable, located in read only memory, results in a Segmentation fault (as expected)
Each class has a pointer to a list of functions, they are each in the same order for derived classes, then the specific functions that are overrided change at that position in the list.
When you point with a base pointer type, the pointed to object still has the correct _vptr.
Base's
Base::function1()
Base::function2()
D1's
D1::function1()
Base::function2()
D2's
Base::function1()
D2::function2()
Further derived drom D1 or D2 will just add their new virtual functions in the list below the 2 current.
When calling a virtual function we just call the corresponding index, function1 will be index 0
So your call
dPtr->function1();
is actually
dPtr->_vptr[0]();
Related
I have a doubt: I can declare a pointer to a class member function
void (*MyClass::myFunc)(void);
and I can declare a pointer to a class member variable
int (MyClass::*var);
My question is: how is an object (composed by member functions and member variables) structured in memory (asm-level) ?
I'm not sure because, except for polymorphism and runtime virtual functions, I can declare a pointer to a member function even without an object and this implies that the code functions are shared among multiple classes (although they require a *this pointer to work properly)
But what about the variables? How come I can declare a pointer to a member variable even without an object instance? Of course I need one to use it, but the fact that I can declare a pointer without an object makes me think a class object in memory represents its variables with pointers to other memory regions.
I'm not sure if I explained properly my doubt, if not just let me know and I'll try to explain it better
Classes are stored in memory quite simply - almost the same way as structures. If you inspect the memory in the place, where the class instance is stored, you'll notice, that its fields are simply packed one after another.
There's a difference though, if your class have virtual methods. In such case the first thing stored in a class instance is a pointer to a virtual method table, which allows virtual methods to work properly. You can read more about this on the Internet, that's a little more advanced topic. Luckily, you don't have to worry about that, compiler does it all for you (I mean, handling VMT, not worrying).
Let's go to the methods. When you see:
void MyClass::myFunc(int i, int j) { }
Actually the compiler converts it into something like:
void myFunc(MyClass * this, int i, int j) { }
And when you call:
myClassInstance->myFunc(1, 2);
Compiler generates the following code:
myFunc(myClassInstance, 1, 2);
Please keep in mind, that this is a simplification - sometimes it's a little more complicated than this (especially when we discuss the virtual method calls), but it shows more or less, how classes are handled by the compiler. If you use some low-level debugger such as WinDbg, you can inspect parameters of the method call and you'll see, that the first parameter is usually a pointer to class instance you called the method on.
Now, all classes of the same type share their methods' binaries (compiled code). Therefore there is no point in making copy of them for each class instance, so there is only one copy held in the memory and all instances use it. It should be clear now, why can you get the pointer to method even if you have no instance of the class.
However, if you want to call the method kept in a variable, you always have to provide a class instance, which can be passed by the hidden "this" parameter.
Edit: In response to comments
You can read more about pointer members in another SO question. I guess, that pointer to member stores the difference between the beginning of classes instance and the specified field. When you try to retrieve the value of a field using the pointer-to-member, compiler locates the beginning of classes instance and move by amount of bytes stored in pointer-to-member to reach the specified field.
Each class instance has its own copy of non-static fields - otherwise they wouldn't be much of a use for us.
Notice, that similarly to pointers to methods, you cannot use pointer to member directly, you again have to provide a class instance.
A proof of what I say would be in order, so here it is:
class C
{
public:
int a;
int b;
};
// Disassembly of fragment of code:
int C::*pointerToA = &C::a;
00DB438C mov dword ptr [pointerToA],0
int C::*pointerToB = &C::b;
00DB4393 mov dword ptr [pointerToB],4
Can you see the values stored in pointerToA and pointerToB? Field a is distant by 0 bytes from the beginning of classes instance, so value 0 is stored in pointerToA. On the other hand, field b is stored after the field a, which is 4 bytes long, so value 4 is stored in pointerToB.
If Base is a base class and Derived a derived class and there are 25 instances of Derived, how are the vtables set up to be accessed by all the instances? Where are they loaded in the memory?
Compilers are allowed to implement dynamic dispatch however they want in c++, i don't think there is actually any requirement to even use a vtable at all, but it would be very unusual to find a compiler that didn't.
In most cases i think that each class (that contains some virtual methods) will own a single vtable (so if i had 5 instances of class A i will still only have 1 vtable), but this behaviour should not be relied upon in any way.
Non virtual classes have no need for vtables as far as i know.
Reading your question it seems as if you think that each object has its own copy of the code, I'm not sure and i don't want to accuse you of anything like that but just in case ...
Google something like: "what does a c++ object look like in memory"
There will be one vtable somewhere in memory, probably in the same place as the code.
Each instance of the class will contain a single pointer to the vtable for that class, so in your case all 25 instances will contain a pointer to one copy of the vtable.
Multiple and virtual inheritance complicate things, but the principle is the same.
I tried to find where exactly virtual function table gets stored for c++ class.
i found some answers like its a "static array of function pointers"
so will it get stored in data segment read only memory? (initialised one)
Most probably yes. However, it's not mandated. It's not even mandated that polymorphism is implemented via virtual function table, but on most platforms it is. These are implementation details, as long as a compiler obeys the behavior set by the standard, it can do whatever it wants.
A vftable is one per class and stored in only one place in memory.
When you make any function virtual, the compiler will insert a vptr inside your class. As a result, the size of the class will grow by 4 bytes (on Win32).This pointer holds the address of the virtual table (vtable). vtable is constructed by the compiler at compile time and is basically nothing but an array of function pointers. The function pointers are actually pointers to the virtual functions of that particular class. To be more exact, the virtual table is a static array of function pointers, so that different instances of the same class can share that vtable. Since, static members are stored in the data section (.data), the vtable is also stored in the data section of the executable.
It's implementation dependent, yes.
And for g++ (4.9.0), virtual table (not the pointer) is stored in the section .rodata of ELF file and its corresponding segment LOAD in memory.
I want to create an instance of a class and place it in shared memory so the same instance can be called from multiple processes. However, this class has virtual methods which I think may cause problems as I have read the mapped data can't contain pointers, which would be the case here with the vtable in the class. Will it work?
As Kerrek SB commented, you cannot map a class containing virtual methods. But you can probably make a simple struct or class without virtuals, map that, and then give a pointer to it to another class which does have virtuals and uses the plain struct as its implementation. Basically, the Pimpl idiom.
If needed, you can even do something like virtual dispatch yourself by storing a "type" integer in the plain struct, and inspecting it to decide which functions to invoke.
Lets say there is a class named Person which contains a virtual function named age(). As per language semantics, vtable is per class and not per object. It is VPTR which is per object and points to vtable.
Questions:
If I build this program (lets say main() exist):
Will the vtable be created i.e can vtable comes in existence w/o even creating a single object?
The address which is put by compiler in vtable for age() is a kind of some static address in memory?
Or is it that compiler internally creates some object for obtaining the address to age() (since age() will be working on some data members which can come in existence only when an object is constructed) or there is some other magic behind this?
As per my understanding, the answers are as follows:
Yes
Yes
Not sure
I tried running "nm" on above program just to see if I can figure out the vtable, but no luck. Is there a way to do so?
Please suggest.
Since it's all implementation defined, my answer describes some 'common implementation'
The v-table is stored in the executable just like the machine-code itself, and loaded to the memory by the OS loader. The OS dosn't care what data to load: string literals, machine code, vtables, constant data, etc...
Suppose you have:
struct A {
int x;
virtual void f() { cout << x; }
};
void g(A* a) { a->f(); }
The generated code will look (semantically) similar to:
// pseudocode, not C++
struct A {
void *vtable;
int x;
};
void A_f(A* this) { cout << this->x; }
void* A_vtable[] = { &A_f };
void g(A* a) { ((void(*)(A*))(((void**)a->vtable)[0]))(a); }
So yes, it's static data.
Of course the above code is very simplified. To support RTTI and virtual inheritance you must do more complicated things.
I don't understand what you mean, but No.
The answers to all of these are completely compiler-dependent. There is no requirement that a physical vtbl even exists; that's just a very common way of implementing the language. But there's no universal ABI for any of this, and it's not something that you as a developer should be worrying about.
This is all compiler dependent, and you should take the whole answer just as a way in which it can be done:
Will the vtable be created i.e can vtable comes in existence w/o even creating a single object?
Depends on the compiler and the program. For example, GCC will create the vtable in the translation unit in which the first virtual function that is not defined inside the class definition is defined (whether any object is created or not), but it might not generate a vtable at all in some cases or it might generate one even if no objects are created.
The address which is put by compiler in vtable for age() is a kind of some static address in memory?
That is usually resolved by the linker/loader. When the program is linked, the linker usually resolves the (relative) addresses of the functions and injects those into the vtable as a first step. When the program is loaded into memory those addresses are fixed to the memory addresses in which the functions are (which depending on the loader can vary from one execution to another).
Or is it that compiler internally creates some object for obtaining the address to age() (since age() will be working on some data members which can come in existence only when an object is constructed) or there is some other magic behind this?
I don't quite follow the whole question. The compiler does not create any object of the type that has the member age artificially, only if your program requests it. The members functions potentially access/modify data members of an object of a given type, but access to those members is handled by means of the implicit this pointer that pass to all non-static members.
"As per language semantics," there's no such thing as a vtable. The C++ specification does not specify exactly how virtual dispatch is implemented. A compiler could use vtables. Or it could use something else; that's up to the compiler.
A specific compiler can use vtables of course. But it can do whatever it wants with them; that's an implementation detail. In short, there's no way for you to know without investigating the specific compiler you're using.
And really, does it matter that much?