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
I've been coding with C for around 10 years and don't think I ever used the keyword "short" before. "unsigned short mINT;" Is this a 16 bit value? If so, why not make it unsigned int mINT; The word "short" to me mentally implies an 8 bit variable. Sorry, Hate to be nitpicking but I prefer code that I understand right off the bat instead of having to pick up my C book (dont have it handy today anyway:) Ditto on checking out unions in C Andy
"The word "short" to me mentally implies an 8 bit variable." That's a bad implication. The Standard specifies that "short" have a minimum range of -2^15 to 2^15.
To be more precise, The Standard specifies that "short" have a minimum range of -32767 to 32767.
"Short" is a standard C type. Like the other C types, the actual width in bits of the type is implementation dependent. The standard tells you that: - a short is at least 16 bits wide - a long is at least 32 bits wide. - an int is at least as wide as a short and no wider than a long. Everything else is up to the implementation. Almost all embedded programmers I know adopt a local convention of typedefs that specify the exact width, e.g. typedef U8 unsigned char; typedef U16 unsigned int; typedef U32 unsigned long; typedef S16 signed int; typedef S32 long; so that they can remain platform independent. The C99 standard introduces a new header, <inttypes.h>, with standard names for these types, e.g. uint8_t uint16_6 The standard also adds some "half open" types with semantics like "any integer width you like as long as it's at least 16 bits wide". Other popular spellings include UINT8, UINT16, etc. You also see BYTE, WORD, DWORD, QWORD. I don't personally care for these because the width of a "word" is really hardware dependent. It's not always 16 bits, though that's what people using this series of names expect. Putting the actual number of bits in the type name is much more clear.
typedef U16 unsigned int; Is only 16 bits in 8 to 16 bit compilers. In 32 bit compilers it will be 32 bits.
typedef U16 unsigned short int; Will give you 16 bits in many 32+ bit compilers. Some compilers only support 32 data.
Thanks for your efforts to help, anyway.