Author: Ezra Hoch
You've probably used classes & interfaces more than once in your delphi programs.
Did you ever dtop to think how delphi implements this creatures ?
Answer:
Inorder to understand this article, you must read the previous article (Inside
Delphi's Classes and Interfaces Part I).
In this article we'll finish covering Delphi's implementation of Interfaces, and
review a few usefull conclusions.
Let's start with an indepth example :
1 type
2
3 IInterface1 = interface
4 procedure ActA;
5 procedure ActB;
6 end;
7
8 IInterface2 = interface(IInterface1)
9 procedure ActC;
10 procedure ActD; stdcall;
11 end;
12
13 TSampleClass = class(TInterfacedObject, IInterface1, IInterface2)
14 procedure ActA;
15 procedure ActB;
16 procedure ActC;
17 procedure ActD; stdcall;
18 end;
19
20 var
21 Interface1: IInterface1;
22 Interface2: IInterface2;
23 Sample: TSampleClass;
24 begin
25 Sample := TSampleClass.Create;
26 Interface1 := Sample;
27 Interface2 := Sample;
28 Interface1.ActA;
29 Interface1.ActB;
30 Interface2.ActA;
31 Interface2.ActB;
32 Interface2.ActC;
33 Interface2.ActD;
34 end;
Instead of looking at the compiled code for this example, I'll simlpy note the
interesting aspects of it. First, when assigning a value to Interface1, we'd expect
delphi to take the value of what 'Sample' points to and add a specific amount
($10) and be done with it. When assigning a value to Interface2, we'd expect delphi
to do the same, just add a smaller amount ($0C) because the interfaces are stored
in memory from the last to the first.
But delphi doesn't do that. It assignes both Interface1 AND Interface2 the value
that 'Sample' points to plus $0C. That's because IInterface2 inherites from
IInterface1. Therefor, IInterface2 includes IInterface1. Hence, any call to
Interface1, will actually be executed through IInterface2's method list.
Second, when we call Interface1.ActA, it calles the 4th (every interface inherites
from IUnknown) method on IInterface2's method list (because IInterface2 inherites
from IInterface1). When we call Interface1.ActB it calles the 5th method on
IInterface2's method list. When we call Interface2.ActA it calles the 4th method on
IInterface2's method list, just the same as Interface1.ActA. That's because
IInterface2 inherites from IInterface1.
Third, when we call Interface2.ActD delphi addes one additional instruction before
calling the 7th method of IInterface2. That's because we've declared a different
convention call to the method (stdcall). Notice that all of IUnknown's methods are
defined with the stdcall directive.
The structor of an interface's method list always follows the following rule :
First Method
.
.
Last Method
The parent's interface's method list
In our case, IInterface2's method list is as follows :
35 ActC
36 ActD
37 // IInterface1's method list
38 ActA
39 ActB
40 // IUnknown's method List
41 QueryInterface
42 _AddRef
43 _Release
NOTE : The structor above is how the methods' code is organized in memory. The
first entry in any interface's method list will belong to QueryInterface (the first
method of IUnknown) but it will point to a place in memory (the implementation of
that specific interface's QueryItnerface method) that is higher than the
interfaces' own methods' implementation - as shown in the structor above. In our
case, IInterface2's QueryInterface's implementation is higher in memory than
IInterface2's ActB's implementation, which is higher in memory than ActD's
implementation. Thou ActD is the 7th entry, ActB is the 5th entry and
QueryInterface is the 1st entry in IInterface2's method list.
To fully understand what happens when delphi calls an interface's method, lets have
a look at the compiled method list of IInterface2 in the example above. The
following code is an exact copy of the compiled code (except for the comments) :
44 // ActC
45 add eax, -$0C
46 jmp TSampleClass.ActC
47 // ActD
48 add dword ptr[esp + $04], -$0C
49 jmp TSampleClass.ActD
50 // ActA
51 add eax, -$0C
52 jmp TSampleClass.ActA
53 // ActB
54 add eax, -$0C
55 jmp TSampleClass.ActB
56 // QueryInterface
57 add dword ptr[esp + $04], -$0C
58 jmp TInterfacedObject.QueryInterface
59 // _AddRef
60 add dword ptr[esp + $04], -$0C
61 jmp TInterfacedObject._AddRef
62 // _Release
63 add dword ptr[esp + $04], -$0C
64 jmp TInterfacedObject._Release
As you remember, an object's method is actually a regular function/procedure that
accepts as a parameter an instance of the method's class. As you can notice, before
each call to the real method ('TSampleClass.ActD' for example) there is one line of
code that changes the value of either 'eax', or 'dword ptr [esp + $04]', depending
on the calling convention. As you can notice, in all cases we subtract $0C form a
variable. But, why 12 ($0C = 12) ? That's because this interface (IInterface2) is
in the 3rd (FRefcount, IUnknown are before it) place after the pointer to VMT of
the clasS TSampleClass. Therefore, the value of any instance of IInterface2 of
TSampleClass (Interface2 for example) is actually the value of the pointer to that
class' instance plus 12.
Here is another example that will help understand the section above. The following
code continues the defenitions from the above code :
65 type
66
67 IAnotherInterface = interface
68 procedure ActE;
69 end;
70
71 TAnotherSample = class(TInterfacedObject, IInterface2, IAnotherInterface)
72 procedure ActA;
73 procedure ActB;
74 procedure ActC;
75 procedure ActD; stdcall;
76 procedure ActE;
77 end;
78
79 var
80 Interface2: IInterface2;
81 begin
82 Interface2 := TAnotherSample.Create;
83 Interface2.ActC;
84 end;
Now, let's compare the entry for this example's IInterface2 and the previous' one :
IInterface2 of TAnotherSample:
add eax, -$10
jmp TAnotherSample.ActC
IInterface2 of TSampleClass:
add eax, -$0C
jmp TSampleClass.ActC
There are two obvious changes :
The actuall function that is called (either TAnotherSample.ActC or
TSampleClass.ActC)
The amount that 'eax' is changed by. Notice that when calling IInterface2 of
TAnotherSample, 'eax' is changed by 16 ($10 = 16) as opposed to being changed by
12. That's because on TAnotherSample, the IInterface2 is the second interface in
the instance's structor in memory, and therefor it is "farther away" from the
instance itself and needs to be changed by additional 4 bytes.
And now to some usefull sutff :
First, if you want to check if 2 (or more) interface variables are of the same
instance, you cannot simply compare them, even if they are of the same type. You
must QueryInterface them to a single interface type, and then compare. As a general
rule, if you want to compare interfaces, QueryInterface them to IUnknown and then
compare.
Example :
85 type
86
87 IBooA = interface
88 end;
89
90 IBooB = interface
91 end;
92
93 TBoo = class(TInterfacedObject, IBooA, IBooB)
94 end;
95
96 var
97 Boo: TBoo;
98 BooA: IBooA;
99 BooB: IBooB;
100 begin
101 Boo := TBoo.create;
102 BooA := Boo;
103 BooB := Boo;
104
105 // This won't complie
106 if BooA = BooB then
107 begin
108 Beep;
109 end;
110
111 if Integer(BooA) = Integer(BooB) then
112 begin
113 // will never get here
114 Beep;
115 end;
116
117 if IUnknown(BooA) = IUnknown(BooB) then
118 begin
119 // will never get here
120 Beep;
121 end;
122
123 // the 'as' word is the same as QueryInterface when acting on interfaces
124 if (BooA as IUnknown) = (BooB as IUnknown) then
125 begin
126 // Will always get here
127 Beep;
128 end;
129 end;
Explaination : The first comparing won't complie, becuase BooA and BooB are of 2
different types. The Second and third comparings will complie but never return
true. That's because type casting doesn't change the value of the variable that's
being type casted. It only allows the complier to complie the code though there are
two different types involved. Hence, if BooA is different from BooB, comparing them
will never return true, no matter what type casting is done to them.
But why do BooA and BooB have different values ? They were both assigned using the
":= Boo;" statment. The answer is simple. Remeber that I said that an interface's
variable's value is actually the value of the instance itself (or at least the
value of the pointer to the instance) plus a different number for each interface ?
In our case, BooA is the same as what Boo points to, added 16. And BooB is the same
as what Boo points to, added 12. That's why BooA and BooB are not that same.
The Forth comparing actually works. That's because if an interface is from the same
type, then comparing it to an interface of that type will always return the
expected result (if both interfaces were aquired via QueryInterface, not by type
casting). That's because if they are of the same type, then the difference between
them and the instance is the same. And if they are of the same instance, then they
must be equal.
That is, each interface is equal to it's instance + a specific Delta (the Delta
depeneds on the interface). In other words, Interface = Instace + Delta. If
'Instance' is the same for both interfaces, and the 'Delta' is the same (cause they
are of the same interface type), then both interfaces must be equal.
Note : This is the way delphi works, for good and for bad. You should take this in
mind when writing code for propertys of interface type. The following code wouldn't
work properly :
130 TSample = class
131 private
132 FData: IUnknown;
133 procedure SetData(Value: IUnknown);
134 protected
135 procedure Changed; virtual; abstract;
136 public
137 property Data: IUnknown read FData write SetData;
138 end;
139
140 procedure TSample.SetData(Value: IUnknown);
141 begin
142 // This is incorrect.
143 if Value <> FData then
144 begin
145 FData := Value;
146 Changed;
147 end;
148 end;
It might seem that this code should work, but it might not work when someone would
assgin the property 'Data' with an IUnknown retreived by a type cast. The correct
code should be :
149
150 procedure TSample.SetData(Value: IUnknown);
151 begin
152 if (Value as IUnknown) <> (FData as IUnknown) then
153 begin
154 FData := Value;
155 Changed;
156 end;
157 end;
Second, each interface you declare that a class implements (with exception of
interfaces that inherite from other interfaces) means that each instance of that
class will take up 4 more byte of memory. That might seem like nothing (and
probably is) except for one case. Consider the following code :
158 IInterfaceA = interface
159 end;
160
161 IInterfaceB = interface
162 end;
163
164 TSampleClass1 = class(TInterfacedObject, IInterfaceA)
165 end;
166
167 TSampleClass2 = class(TSampleClass1, IInterfaceA, IInterfaceB)
168 end;
It would seem that each instance of TSampleClass1 should take up 16 bytes, and each
instance of TSampleClass2 should take up 20 bytes (4 bytes more, because it
supports one more interface). That is not true. Each instance of TSampleClass1 does
take up 16 byte. But, each instance of TSampleClass2 takes up 24 bytes ! That's
because delphi creates an interface entry even for interfaces that are already
implemented by parent classes.
The solution to this is simple. Just remove the decleration of IInterfaceA from
TSampleClass2. This will not change the fact that TSampleClass2 implements
IInterfaceA, cause TSamlpeClass2 inherites from TSamlpeClass1, which implements
IInterface1. This wouldn't have happened if IInterfaceB was a decendant of
IInterfaceA.
This might add up to quit alot if you do your inheritence improporely. For example
:
169 TSampleClass1 = class(TInterfacedObject, IUnknown)
170 end;
171
172 TSampleClass2 = class(TSampleClass1, IUnknown)
173 end;
174
175 TSampleClass3 = class(TSampleClass2, IUnknown)
176 end;
177
178 TSampleClass4 = class(TSampleClass3, IUnknown)
179 end;
180
181 TSampleClass5 = class(TSampleClass4, IUnknown)
182 end;
Each instance of TSampleClass5 takes up 32 bytes of memory, though it has no real data (except for FRefCount of TItnerfacedObject).
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