One of the main reasons why I don’t like object oriented languages is because of the inefficiencies the language usually introduces on the runtime code. If you’ve ever traced through a method dispatch routine then you know what I mean (what ever happened to the days of simple, direct JSR’s?). This month you have a chance to write a fast method dispatcher. Who knows? If it’s efficient enough I might just toss my assembler and use your dispatcher with a high level language instead...
The prototype of the function you write is:
typedef unsigned short ushort;
typedef ushort ClassID;
typedef ushort MethodNumber;
MethodAddress
FindMethod(theClassID, theMethodNumber)
ClassID theClassID;
MethodNumber theMethodNumber;
TheMethodNumber is the number of the method you’re trying to find the address of and theClassID is the ID of the class you want it for. You’ll pass theClassID to a function called GetClassPtr to get a pointer to a Class data structure, which looks like this:
typedef void *MethodAddress;
typedef struct {
MethodNumber methodNumber;
MethodAddress methodAddress;
} MethodEntry;
typedef struct {
ushort inheritedCount;
ushort inheritedClasses[15];
MethodNumber largestMethodNumber;
ushort methodCount;
MethodEntry methods[];
} Class, *ClassPtr;
The function GetClassPtr will be part of my test bench, although you’ll have to implement at least a rudimentary version of it to test your program (or you can e-mail me for a sample version):
ClassPtr
GetClassPtr(classID)
unsigned short classID;
If GetClassPtr returns -1 (kClassNotFound) then the class cannot be found and your FindMethod routine should return 0 (kMethodNotFound). I will be providing sample data and a sample GetClassPtr function for those who are interested. To get a copy, send me e-mail at scanlin@genmagic.com (internet) or any of the Programmer Challenge addresses listed on page 2.
Once you have a ClassPtr you should look in that class’s methods[] array to see if you can find an entry whose methodNumber is equal to theMethodNumber. Methods[] is a variable-length array (thus, making Class a variable-size structure) containing methodCount number of entries which are sorted smallest to largest by methodNumber. MethodCount is 1-based and is always greater than zero. If you find a match then you should return the corresponding methodAddress.
If you don’t find a match then you should look at the inherited classes (starting with index zero) to see if the method is implemented by one of this class’s superclasses. We support multiple inheritance here and the number of classes we inherit from is stored in inheritedCount (which will be from zero to 15). The class IDs of the classes we inherit from are stored in the inheritedClasses[ ] array. You can pass any of the entries in inheritedClasses to GetClassPtr to get a ClassPtr to that class.
If you can’t find the requested methodNumber in any part of the inheritance tree then FindMethod should return zero (kMethodNotFound).
Here’s a simple example. These 54 bytes (starting at location 0x1000) represent class ID 5:
1000:00000000 00000000 00000000 00000000
1010:00000000 00000000 00000000 00000000
1020:00680003 0023AAAA AAAA0057 BBBBBBBB
1030:0068CCCC CCCC
The short at location 1000 (inheritedCount) tells us that there are no inherited classes for this class. The short at location 1022 (methodCount) tells us that this class has 3 methods. Methods[0] is from 1024 to 1029; the methodNumber is 23 and the methodAddress is AAAAAAAA (this is just test data to illustrate the structure). Methods[1] is from 102A to 102F and methods[2] is from 1030 to 1035. The short at location 1020 (largestMethodNumber) is equal to the methodNumber of the last MethodEntry in the list (which is the largest methodNumber overall since the list is sorted). In other words, the expression theClassPtr->largestMethodNumber == theClassPtr-> methods[theClassPtr->methodCount-1].methodNumber is always true.
If this class had inherited from class 7 and class 9 then it would have looked like this instead:
1000:00020007 00090000 00000000 00000000
1010:00000000 00000000 00000000 00000000
1020:00680003 0023AAAA AAAA0057 BBBBBBBB
1030:0068CCCC CCCC
In either case, if you call GetClassPtr(5), since this is class 5 we’re looking at, you would have the value 0x1000 (as type ClassPtr) returned to you.
Since I’ll be calling FindMethod several thousand times with the same set of classes (just like a real runtime system!) you’ll probably want to implement some kind of cache. And since it is desirable for runtime systems to take as little memory as possible, we’re going to have a rule that says your code cannot use more than 16K of memory for its cache (use a static to keep a pointer to it). The total number of methods in the set of classes I’ll be testing with is about 5000, numbered from 1 to 5000. The total number of classes is about 400, numbered from 1 to 400. Those 400 classes will implement an average of 15 methods each and will inherit from an average of 5 other classes (that’s 5 total, once you’ve walked the entire inheritance tree for a particular class). Of course, some methods will be called frequently while others are hardly ever called.
Because this is a little complex, I’m going to give you the brute force way of doing what I’ve described. I’m sure you can do better than this (I’ve used short variable names so that the code will fit in the magazine column):
MethodAddress
FindMethod(cid, mn)
ClassID cid;
MethodNumber mn;
{
ClassPtr cp;
MethodAddress addr;
int i;
cp = GetClassPtr(cid);
if (cp == kClassNotFound)
return kMethodNotFound;
/* look in this class */
i = 0;
do {
if (mn == cp->methods[i].methodNumber)
return cp->methods[i].methodAddress;
i++;
} while (i < cp->methodCount);
/* look in superclasses */
i = 0;
while (i < cp->inheritedCount) {
addr = FindMethod( cp->inheritedClasses[i], mn);
if (addr != kMethodNotFound)
return addr;
i++;
}
return kMethodNotFound;
}
E-mail me if you have any questions or if you want the sample data and GetClassPtr function. And if you want to see your name in print all you have to do is either enter a challenge or have me use one of your suggested challenges.
