Hi all, I'm new to C and need a hint on an optimization problem. I have a 16 bit counter, divided into two 8bit SFRs, let's say TH0 and TL0, and I want to transfer them to a short. I could: mshort = (short)TH0 * 256 + TL0 but this takes 16bytes, similar to something like: mshort = (short)TH0 << 8 + TL0 I would like to do it more simple, like: MSB of mshort = TH0 LSB of mshort = TL0 I tried it that way: unsigned short *pINT; pINT = &TH0; but this is not possible with SFRs. Something else I tried: unsigned short mINT; unsigned char *pChar; pChar = & mINT; *pChar = TH0; *(pChar + 1) = TL0; results in even more code (31 bytes) So I think it would be the easiest, if I could place 2 char variables at the same address, covered by my short, like: unsigned short mINT; unsigned char mChar[2] (at the same address like mINT) mChar[0] = TH0; mchar[1] = TL0; but I don't know, how to place the mChar array at the same address of mINT. I tried it with: unsigned char mChar[2] _at_ mINT; but this is not possible (compiler error). So anyone here, who could give me a hint how to place both variables at the same address? Thanks in advance Marco Della Rocca
if the two SFRs are placed in contiguous addresses, you can just define an appropriate SFR16, a very dangerous practice. Two years from now you will use another derivative where they are not "in contiguous addresses" and you will spend hours/days/weeks figuring out why this 'good' code suddenly does not work. This seems to be, based on reading some past posts, one of the bugs that are most difficult to catch for some. With an ICE it is no big deal. Erik PS in my opinion, SFR16 should be outlawed (it is here) for the above reason. Erik