How to parse uint16_t to int in C?

Question

I got the data for uint16_t and it like 0x2c06.

For the 0x2c06 , 2c is the first data and the 06 is the second data.

How do I parse the 0x2c06 and turn the 2c and the 06 to int?

Thanks in advance.

It is not clear what you are asking. Is your input a string/char* or a `uint16` variable? Also `0x2c` is called a **Byte** — DrKoch, May 04 '15 at 07:23

score 1 · Answer 1 · edited Dec 17 '20 at 22:32

1

You can use a char* and some pointer arithmetic to get this done.

Pseudo-code:

uint16_t var = 0x2c06;
unsigned char * temp = &var;

int first = *temp;               //or second, see note
int second = *(temp + 1);       // or first, see note

_{Note: You need to take care of endianness}

edited Dec 17 '20 at 22:32

Kyrol

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answered May 04 '15 at 07:19

Sourav Ghosh

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The result of this depends on the endianess of the host cpu. – trenki May 04 '15 at 07:26
@trenki yeah, absolutely. That needs to be taken care of. I'll adda line about it in my answer itself. – Sourav Ghosh May 04 '15 at 07:28
1

If you use `char`, then signedness is implementation defined. Better to use `unsigned char` to avoid problems. – user694733 May 04 '15 at 07:50

phil · Answer 2 · 2015-05-04T07:36:00.383

-2

Build a set of case operators cast the uint and just equate it, eg

switch(sizeof unsign) 

{

case sizeof (unsigned short int):

  int result = (unsigned short int)unsign;

  break;

default:

  fprintf (stderr, "recast of unit_16  failed.\n");

}

Sorry if there's bugs in this, it's just typed into the browser.

edited May 04 '15 at 07:36

answered May 04 '15 at 07:28

phil

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How to parse uint16_t to int in C?

2 Answers2