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
Hi, if the two SFRs are placed in contiguous addresses, you can just define an appropriate SFR16, e.g.:
sfr TL2 = 0xCC; sfr TH2 = 0xCD; sfr16 T2 = 0xCC;
you can just define an appropriate SFR16, Bad idea. Never define your own SFRs. That's what you get ready-made headers from Keil for.
That's what you get ready-made headers from Keil for. Which are sometimes woefully incomplete and/or just copied from a completely different device without modification *cough*ADuC845*cough*.
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