read two bytes at once from a byte array in C

Question

Let's say I have the following array of bytes:

uint8_t barr[4] = {0xCE, 0xCE, 0xCE, 0xCE};

Given an index n, I want to be able to read two bytes:

uint16_t d16 = barr[0];

And have d16 be equal to

0xCECE

Perhaps there is a function in the standard library that can perform such task?

All of your sample values are the same. Should 0x01, 0x02 be paired as 0x0102 or as 0x0201? Do you know the endianness of the data and of your own system? — Jongware
– Jongware, Commented Dec 27, 2017 at 13:49

iBug · Accepted Answer · 2017-12-27 16:06:45Z

9

A piece of cake:

memcpy(&d16, barr + n, sizeof(d16));

Don't try to convert pointers or use unions. Those either are undefined behaviors, or may trip trap representations. memcpy() is the "canonical" solution (as the C++ boost library does).

edited Dec 27, 2017 at 16:06

answered Dec 27, 2017 at 13:42

iBug

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6 Comments

Gnqz Over a year ago

All wise, a simple and sufficient answer.

Jacek Cz Over a year ago

@JanParzydło Not so easy, propably You should read about endianess. Think sample with different values { 0x11, 0x12, 0x13, 0x14 }

alk Over a year ago

Why is boost a reference when doing C?

iBug Over a year ago

@alk This is the same case across C and C++.

iBug Over a year ago

@alk Fine. I have to agree.

|

alk · Accepted Answer · 2017-12-28 08:09:49Z

0

A linear and endianness-safe approach here:

#include <stdint.h>
#include <stdio.h>
#include <arpa/inet.h> /* for ntohs() */

union Convert
{
  uint16_t  to_integer_big_endian;
  uint8_t barr[sizeof (uint16_t)];
};

uint16_t pick_and_convert(uint8_t * barr, size_t s, size_t n)
{
  union Convert c = {0};  /* Take out-of-bound bytes as 0s. */

  for (size_t i = 0; i < sizeof c.barr && (n + i) < s; ++i)
  {
    c.barr[i] = barr[n + i];
  }

  return ntohs(c.to_integer_big_endian);
}

int main(void)
{
   uint8_t barr[4] = {1, 2, 3, 4}; /* Assuming a binary sequence. */
   size_t n;

   scanf("%zu", &n);

   uint16_t to_integer = pick_and_convert(barr, sizeof barr / sizeof *barr, n);

   printf("result = 0x%hx\n", to_integer);
 }

edited Dec 28, 2017 at 8:09

answered Dec 27, 2017 at 16:57

alk

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