Conversion of uchar array to uint8 type pointer

Question

I want to convert a array of uchar to uint8 pointer. As both are of 8 bits and value ranges from 0 to 255 so I do not think it should cause and issue.

uchar list[100];

I have to pass above list to a function which accepts pointer to uint8t. Can i pass it like this :

(uint8t *)list

such that it will not affect the value when i read it as uint8.

Yes that's fine and correct, if you are talking about the standard type `uint8_t`. — M.M, Mar 23 '14 at 05:53
Well, if you mean `unsigned char` , and unicode char, it should be OK usually. The standard says char should be at least 8 bits wide, to allow C compilers on machines where treating 9, 10, 12, 16 etc. bits as char is more simple, but my guess is probably you are not going run this on such machines. — Gábor Buella, Mar 23 '14 at 06:27
i tried, it works fine just i wanted to be sure that it doesn't break in extreme or borderline cases. — Prannoy Mittal, Mar 23 '14 at 09:39

score 0 · Answer 1 · answered Mar 23 '14 at 13:16

Well, uchar and uint8_t are not necessarily the same width ! Technically, a character can be 16bit wide. It is a rare, but it happens, especially in embedded world. So uchar may be not an 8bit type, but uint8 is always a 8bit type.

If uchar is indeed not a byte, then casting from uchar []to uint8_t * eliminates alignment restrictions. For platforms that do not support miss-aligned access (smells like embedded again!) trying to cast like this would be a bug. If you compile with GCC, add -Wcast-align flag to catch problems like this.

Conversion of uchar array to uint8 type pointer

1 Answers1