Any documentaion available for behaviour of unsigned char pointer in C?

Question

What is an unsigned char pointer and how is it different from a char pointer in C?

Not much information on unsigned char pointer in C , and How to print the data pointed by an unsigned char pointer?

Any explanation on this will be helpful.

The problem is in the line of code you didn't show. Otherwise the message is pretty clear: %d expcects an int and you provide an unsigned int*. This is _undefined behaviour_ (google that) — Jabberwocky, Mar 18 '20 at 11:03
Just added a sample c code snippet : unsigned char* temp; temp= 133+2; printf("%d",temp); — Vibhuti Sawant, Mar 18 '20 at 11:09
You probably meant to do `printf("%u",(unsigned int)*temp);` — Lundin, Mar 18 '20 at 11:11
I think you should read the chapter dealing with pointers in your C text book. `char* temp; temp= 133+2;` is pretty pointless, maybe you meant `char temp; temp= 133+2;`? — Jabberwocky, Mar 18 '20 at 11:12
@Lundin I don't think so, look at his code snippet in the second comment — Jabberwocky, Mar 18 '20 at 11:13
@P__J__ probably, but without more information from the OP it's hard to tell. — Jabberwocky, Mar 18 '20 at 11:16
@VibhutiSawant what is you code supposed to do? What output do you expect. Probably just `unsigned char* temp;` -> `unsigned char temp;`, but we need more information — Jabberwocky, Mar 18 '20 at 11:25
@Jabberwocky the main aim to ask the question was to know the functionality of unsigned char pointer, and how to get the data its pointing to. I am more interested in knowing the memory representation and the byte of memory temp is referring to. — Vibhuti Sawant, Mar 18 '20 at 11:57

score 1 · Answer 1 · answered Mar 18 '20 at 11:09

1

Assuming you have something like

unsigned char *p = something;

then

printf( “p = %p\n”, (void *)p );

prints the value of the pointer itself, while

printf( “*p = %hhu\n”, *p );

prints the value of what p points to as a decimal integer (unsigned).

answered Mar 18 '20 at 11:09

John Bode

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Not quite sure if this is the correct answer. I suspect the OP doesn't even know what a pointer is, but without more information it's hard to tell – Jabberwocky Mar 18 '20 at 11:18

0___________ · Answer 2 · 2020-03-18T11:09:13.937

0

the documentation is here: https://linux.die.net/man/3/printf

unsigned char x = 0x4d;
unsigned char *y = &x;

printf("pointer: %p, %hhu\n", (void *)y, *y);

endianess do not matter as sizeof(unsgined char) id always 1.

If you want to pun types via pointers - it is generally a bad idea.

edited Mar 18 '20 at 11:09

answered Mar 18 '20 at 11:06

0___________

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Not quite sure if this is the correct answer, apparently the OP provided the a pointer instead of value, but without seeing his code it's hard to tell. – Jabberwocky Mar 18 '20 at 11:07

score 0 · Answer 3 · answered Mar 18 '20 at 11:16

How to print the data pointed to by an unsigned char* p?

That depends entirely on the datas format. And what you want actually have and what you need.

Is it just a binary blob of known length?
You probably want a hexdump then. Search for it, while it is common, it is not builtin, and there are many options.

Is it really a null terminated string?
Print it using %s after casting to char*.

Is it just a single unsigned char?
Print it using %hhu after dereferencing.

There is really insufficient information to judge what is needed.

Any documentaion available for behaviour of unsigned char pointer in C?

3 Answers3