问题
I'm reading Bjarne's paper: Multiple Inheritance for C++.
In section 3, page 370, Bjarne said that "The compiler turns a call of a member function into an "ordinary" function call with an "extra" argument; that "extra" argument is a pointer to the object for which the member function is called."
I'm confused by the extra this argument. Please see the following two examples:
Example 1:(page 372)
class A {
int a;
virtual void f(int);
virtual void g(int);
virtual void h(int);
};
class B : A {int b; void g(int); };
class C : B {int c; void h(int); };
A class c object C looks like:
C:
----------- vtbl:
+0: vptr --------------> -----------
+4: a +0: A::f
+8: b +4: B::g
+12: c +8: C::h
----------- -----------
A call to a virtual function is transformed into an indirect call by the compiler. For example,
C* pc;
pc->g(2)
becomes something like:
(*(pc->vptr[1]))(pc, 2)
The Bjarne's paper told me the above conclusion. The passing this point is C*.
In the following example, Bjarne told another story which totally confused me!
Example 2:(page 373)
Given two classes
class A {...};
class B {...};
class C: A, B {...};
An object of class C can be laid out as a contiguous object like this:
pc--> -----------
A part
B:bf's this--> -----------
B part
-----------
C part
-----------
Calling a member function of B given a C*:
C* pc;
pc->bf(2); //assume that bf is a member of B and that C has no member named bf.
Bjarne wrote: "Naturally, B::bf() expects a B* (to become its this pointer)." The compiler transforms the call into:
bf__F1B((B*)((char*)pc+delta(B)), 2);
Why here we need a B* pointer to be the this?
If we just pass a *C pointer as the this, we can still access the members of B correctly I think. For example, to get the member of class B inside B::bf(), we just need to do something like: *(this+offset). this offset can be known by the compiler. Is this Right?
Follow up questions for example 1 and 2:
(1) When it's a linear chain derivation (example 1), why the C object can be expected to be at the same address as the B and in turn A sub-objects? There is no problem to use a C* pointer to access class B's members inside the function B::g in example 1? For example, we want to access the member b, what will happen in runtime? *(pc+8)?
(2) Why can we use the same memory layout (linear chain derivation) for the multiple-inheritance? Assuming in example 2, class A, B, C have exactly the same members as the example 1. A: int a and f; B: int b and bf (or call it g); C: int c and h. Why not just use the memory layout like:
-----------
+0: a
+4: b
+8: c
-----------
(3) I've wrote some simple code to test the differences between the linear chain derivation and multiple-inheritance.
class A {...};
class B : A {...};
class C: B {...};
C* pc = new C();
B* pb = NULL;
pb = (B*)pc;
A* pa = NULL;
pa = (A*)pc;
cout << pc << pb << pa
It shows that pa, pb and pc have the same address.
class A {...};
class B {...};
class C: A, B {...};
C* pc = new C();
B* pb = NULL;
pb = (B*)pc;
A* pa = NULL;
pa = (A*)pc;
Now, pc and pa have the same address, while pb is some offset to pa and pc.
Why the compile make these differences?
Example 3:(page 377)
class A {virtual void f();};
class B {virtual void f(); virtual void g();};
class C: A, B {void f();};
A* pa = new C;
B* pb = new C;
C* pc = new C;
pa->f();
pb->f();
pc->f();
pc->g()
(1) The first question is about pc->g() which relates to the discussion in example 2. Does the compile do the following transformation:
pc->g() ==> g__F1B((*B)((char*)pc+delta(B)))
Or we have to wait for the runtime to do this?
(2) Bjarne wrote: On entry to C::f, the this pointer must point to the beginning of the C object (and not to the B part). However, it is not in general known at compile time that the B pointed to by pb is part of a C so the compiler cannot subtract the constant delta(B).
Why we cannot know the B object pointed to by pb is part of a C at the compile time? Based on my understanding, B* pb = new C, pb points to a created C object and C inherits from B, so a B pointer pb points to part of C.
(3) Assume that we do not know B pointer to by pb is part of a C at the compile time. So we have to store the delta(B) for the runtime which is actually stored with the vtbl. So the vtbl entry now looks like:
struct vtbl_entry {
void (*fct)();
int delta;
}
Bjarne wrote:
pb->f() // call of C::f:
register vtbl_entry* vt = &pb->vtbl[index(f)];
(*vt->fct)((B*)((char*)pb+vt->delta)) //vt->delta is a negative number I guess
I'm totally confused here. Why (B*) not a (C*) in (*vt->fct)((B*)((char*)pb+vt->delta))???? Based on my understanding and Bjarne's introduction at the first sentence at 5.1 section an 377 page, we should pass a C* as this here!!!!!!
Followed by the above code snippet, Bjarne continued writing: Note that the object pointer may have to be adjusted to po int to the correct sub-object before looking for the member pointing to the vtbl.
Oh, Man!!! I totally have no idea of what Bjarne tried to say? Can you help me explain it?
回答1:
Bjarne wrote: "Naturally, B::bf() expects a B* (to become its this pointer)." The compiler transforms the call into:
bf__F1B((B*)((char*)pc+delta(B)), 2);
Why here we need a B* pointer to be the this?
Consider B in isolation: the compiler needs to be able to compile code ala B::bf(B* this). It doesn't know what classes might be further derived from B (and the introduction of derived code might not happen until long after B::bf is compiled). The code for B::bf won't magically know how to transform a pointer from some other type (e.g. C*) to a B* it can use to access data members and RunTime Type Info (RTTI / virtual dispatch table, typeinfo).
Instead, the caller has the responsibility of extracting a valid B* to the B sub-object in whatever actual runtime type is involved (e.g. C). In this case, the C* holds the address of the start of the overall C object which likely matches the address of the A sub-object, and the B sub-object is some fixed but non-0 offset further into memory: it's that offset (in bytes) that must be added to the C* in order to get a valid B* with which to call B::bf - that adjustment is done when the pointer is cast from C* type to B* type.
(1) When it's a linear chain derivation (example 1), why the C object can be expected to be at the same address as the B and in turn A sub-objects? There is no problem to use a
C*pointer to access class B's members inside the functionB::gin example 1? For example, we want to access the member b, what will happen in runtime?*(pc+8)?
Linear derivation B : A and C : B can be thought of as successively tacking B-specific fileds on the end of A, then C-specific fields on the end of B (which is still B-specific fields tacked on the end of A). So the whole thing looks like:
[[[A fields...]B-specific-fields....]C-specific-fields...]
^
|--- A, B & C all start at the same address
Then, when we talk about a "B" we're talking about all the embedded A fields as well as the additions, and for "C" there's still all the A and B fields: they all start at the same address.
Regarding *(pc+8) - that's right (given the understanding that we're adding 8 bytes to the address, and not the usual C++ behaviour of adding multiples of the pointee's size).
(2) Why can we use the same memory layout (linear chain derivation) for the multiple-inheritance? Assuming in example 2, class A, B, C have exactly the same members as the example 1. A: int a and f; B: int b and bf (or call it g); C: int c and h. Why not just use the memory layout like:
-----------
+0: a
+4: b
+8: c
-----------
No reason - that's exactly what happens... the same memory layout. The difference is that the B subobject doesn't consider A to be a part of itself. It's now like this:
[[A fields...][B fields....]C-specific-fields...]
^ ^
\ A&C start \ B starts
So when you call B::bf it wants to know where the B object starts - the this pointer you provide should be at "+4" in the above list; if you call B::bf using a C* then the compiler-generated calling code will need to add that 4 in to form the implicit this paramter to B::bf(). B::bf() can't simply be told where A or C start at +0: B::bf() knows nothing about either of those classes and doesn't know how to reach b or its RTTI if you give it a pointer to anything other than its own +4 address.
回答2:
Function bf() in your example is a member of class B. Inside B::bf() you will be able to access all members of B. That access is performed through this pointer. So in order for that access to work properly, you need this inside B::bf() to point precisely to B. This is why.
The implementation of B::bf() does not know whether this B object is a standalone B object, or a B object embedded into C object, or some other B object embedded into something else. For that reason, B::bf() cannot perform any pointer corrections for this. B::bf() expects all pointer corrections to be done in advance, so that when B::bf() begins execution, this points precisely to B and nowhere else.
This means that when you are calling pc->bf(), you have to adjust the value of pc by some fixed offset (offset of B in C) and use the resultant value as this pointer for bf().
回答3:
Maybe this makes more sense if you ignore the function call for now and instead consider the conversion of a C* to a B* that is required before calling bf(). Since the B subobject doesn't start at the same address as the C object, the address needs to be adjusted. In cases where you have only one baseclass, the same is done, but the offset (delta(B)) is zero, so it is optimized out. Then, only the type attached to the address is changed.
BTW: Your quoted code (*((*pc)[1]))(pc, 2) doesn't perform this conversion, which is formally wrong. Since it isn't real code anyway, you have to infer that by reading between the lines. Maybe Bjarne just intended to use the implicit conversion to baseclass there.
BTW 2: I think you misunderstand the layout of classes with virtual functions. Also, just as disclaimer, the actual layouts depend on the system, i.e. compiler and CPU. Anyhow, consider two classes A and B with a single virtual function:
class A {
virtual void fa();
int a;
};
class B {
virtual void fb();
int b;
};
The layout would then be:
----------- ---vtbl---
+0: vptr --------------> +0: A::fa
+4: a ----------
-----------
and
----------- ---vtbl---
+0: vptr --------------> +0: B::fb
+4: b ----------
-----------
In words, there are three guarantees for class A (those for B are equivalent):
- Given a pointer
A*, at offset zero to that pointer I find the address of the vtable. At position zero of that table, I find the address of the functionfa()for that object. While the actual function may change in derived classes (due to overrides), the offset in the table is fixed. - The type of the function in the vtable is fixed, too. At position zero of the vtable is a function that takes a hidden
A* thisas parameter. The actual function may be overridden in a derived class, but the type of the function here must be retained. - Given a pointer
A*, at offset four to that pointer I find the value of the member variablea.
Now, consider a third class C:
class C: A, B {
int c;
virtual void fa();
};
Its layout would be like
----------- ---vtbl---
+0: vptr1 -------------> +0: A::fa
+4: a
+8: vptr2 -------------> +4: B::fb
+12: b +8: C::fc
+16: c ----------
-----------
Yes, this class contains two vtable pointers! The reason is simple: The layout of classes A and B is fixed when they are compiled, see above guarantees. In order to allow substituting a C for an A or B (Liskov Substitution Principle), these layout guarantees must be retained, since the code handling the object only knows about e.g. A, but not C.
Some remarks on this:
- Above, you already find an optimization, the vtable pointer for class
Chas been merged with the that for classA. This simplification is only possible for one of the baseclasses, hence the difference between single and multiple inheritance. - When calling
fb()on an object of typeC, the compiler must callB::fbwith a pointer so that the guarantees above are met. For that, it has to adjust the address of the object so that it points to aB(offset +8) before calling the function. - If
Coverridesfb(), the compiler will generate two versions of that function. One version is for the vtable of theBsubobject, which then takes anB* thisas hidden parameter. The other will be for the separate entry in the vtable of theCclass and it takes aC*. The first one will only adjust the pointer from theBsubobject to theCobject (offset -8) and call the second one. - The above three guarantees are not necessary. You could also store the offset of the member variables
aandbinside the vtable. Similarly, the adjustment of the address during function call could be done indirectly via information embedded inside the object via its vtable. This would be much less efficient though.
回答4:
It should in theory be that the compiler would take any this's in the code and if refer to the pointer so it knows what the this is referring to.
来源：https://stackoverflow.com/questions/30747633/compilers-detail-of-this-pointer-virtual-function-and-multiple-inheritance