Re: Pointer arithmetic error

[Bernd Petrovitsch <bernd@firmix.at>]

On Fre, 2008-06-27 at 11:55 +0100, David Given wrote:
> Bernd Petrovitsch wrote:
> [...]
> > In C, there is no type "byte" (unless you typedef oder #define it).
> > "byte" is usually (but not necessarily) meant as "unsigned char".
> 
> The issue here is that sparse is helpful and converts pointer offsets 
> into byte offsets when generating code for pointer arithmetic. So:
> 
> const int* p = (const char*) 1234;
> p += 10;
> 
> -->
> 
> set.32 %p0 <- 1234
> add.32 %p1 <- %p0, 40
> (In fact, in this example it'd collapse these together and use 1274 
> instead.)
> 
> This appears to be done using hard-coded knowledge that a byte (one unit 
> of addressingness) is 8 bits wide.

It is done IMHO with the false knowledge that "sizeof(int) == 4 *
sizeof(char)".
IIUC your hardware (a DSP or what is it exactly) has "sizeof(int) ==
sizeof(char)" so the size of a char (or int) as such is irrelevant.

> Chris Li wrote:
> > Byte need to big enough to hold the char. Using bits_in_byte is better.
> > There might be other place in sparse assume byte is 8 bits.
> 
> IIRC C specifies that sizeof() returns values measured in chars, but I 

ACK. Therefore "sizeof(char) == 1" must always hold.

> don't believe it specifies any mapping between the size of chars and the 
> underlying addressing units --- it should be possible to use 16-bit 

Yes, that's what CHAR_BIT is for.

> chars, for example, on an 8-bit byte system. Using 32-bit ints, 
> sizeof(int) would then return 2; but you wouldn't be able to access 
> individual bytes from C.

ACK (apart from "shift and mask it" ans similar).

	Bernd
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