Question

There is a class with three templates.

#if defined(USE_CACHE_FALRU)
template class Cache<FALRU>;
#endif

#if defined(USE_CACHE_IIC)
template class Cache<IIC>;
#endif

#if defined(USE_CACHE_LRU)
template class Cache<LRU>;
#endif

These template has common function:

FALRUBlk* accessBlock(Addr addr, int &lat, int context_src, int *inCache = 0);
IICTag* accessBlock(Addr addr, int &lat, int context_src);
BlkType* accessBlock(Addr addr, int &lat, int context_src);

As you can see, number of arguments for one of the templates differs from others.

Now in Cache(), there is a function which call accessBlock()

template<class TagStore>
bool Cache<TagStore>::access(PacketPtr pkt, BlkType *&blk, int &lat, PacketList &writebacks)
{ 
 ...
 blk = tags->accessBlock(pkt->getAddr(), lat, id);
 ...
}

In a confing file, all templates are defined

#define USE_CACHE_LRU 1
#define USE_CACHE_FALRU 1
#define USE_CACHE_IIC 1 

I wonder how this file is compiled. As you can see, FALRUBlk::accessBlock() takes 4 arguments. However in Cache::access(), only three arguments are passed.
So Can someone explain how this function compiled without problem?

Answer 1

It compiles fine, because the default value for the 4th is used :

FALRUBlk* accessBlock(Addr addr, int &lat, int context_src, int *inCache = 0);

Other 2 have only 3 arguments :

IICTag* accessBlock(Addr addr, int &lat, int context_src);
BlkType* accessBlock(Addr addr, int &lat, int context_src);

Therefore, you can always call that method like in your example :

blk = tags->accessBlock(pkt->getAddr(), lat, id);