Convert unsigned char array into uint8_t array?

Question

I need convert unsigned char array into unit8_t array maybe someone has an idea how to do it?

I tried to look for information on the Internet, but unfortunately I did not succeed.

:)

There's nothing to convert. **Iff** `uint8_t` exists, it is the same as `unsigned char`. Or the other way around: If `unsigned char` has more than 8 bits, there's no `uint8_t` available. — , Jul 28 '17 at 11:01

score 6 · Accepted Answer · answered Jul 28 '17 at 10:59

6

So, have you tried uint8_t* myuint8array = (uint8_t*)myuchararray; ?

answered Jul 28 '17 at 10:59

Mike Nakis

score 1 · Answer 2 · answered Jul 28 '17 at 11:02

1

You have

unsigned char arr[size];

and you want

uint8_t arr[size];

uint8_t is simply defined as

typedef unsigned char uint8_t;

:-)

answered Jul 28 '17 at 11:02

Robert

1

Or not defined at all, in case `char` has more than 8 bits. – Jul 28 '17 at 11:02
And when would a character possibly have more than 8 bits? – Robert Jul 28 '17 at 11:03
@Robert - It could for example be 9 bits [on a 36-bit computer](https://stackoverflow.com/a/6972551/597607). – Bo Persson Jul 28 '17 at 11:05
character != `char`. And then, the answer is "*when it has more than 8 bits*". Yes, such platforms exist and they can't provide `uint8_t` for obvious reasons. – Jul 28 '17 at 11:05
This was complete news to me, i have never heard of this before. Thanks, will read this straight away. :-) – Robert Jul 28 '17 at 11:07
@RobertK because such platforms exist, C only mandates a *minimum* of 8 bits for `char`. But `uint8_t` must have **exactly** 8 bits, so this type will be missing on a platform having larger `char`s. – Jul 28 '17 at 11:10
@RobertK that is why `#define CHAR_BIT 8` (for example) appears in `limits.h` – Weather Vane Jul 28 '17 at 11:20
@WeatherVane Interesting. Learn something new every day :-) – Robert Jul 28 '17 at 11:31
@RobertK funny you should say that ... so do I! – Weather Vane Jul 28 '17 at 11:42

2 Answers2