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 -- Firmix Software GmbH http://www.firmix.at/ mobil: +43 664 4416156 fax: +43 1 7890849-55 Embedded Linux Development and Services -- To unsubscribe from this list: send the line "unsubscribe linux-sparse" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
