convert int to char array Big Endian

Question

I found this code on SO.

unsigned char * SerializeInt(unsigned char *buffer, int value)
{
  /* Write big-endian int value into buffer; assumes 32-bit int and 8-bit char. */
  buffer[0] = value >> 24;
  buffer[1] = value >> 16;
  buffer[2] = value >> 8;
  buffer[3] = value;
  return buffer + 4;
}

You can see the code claims it is writing an integer into the buffer, in a Big Endian Way.

My question is: Does this function work correctly on both Little endian and Big endian machines? In other words, on both LE and BE machines, does it store the number to buffer in a Big Endian Way? And if yes, Why?

@leeduhem: Aha, that was my point. That is why I asked I am not sure either.. But I started to have doubts that it may work on both BE and LE machines — , Mar 14 '14 at 09:14

score 5 · Accepted Answer · answered Mar 14 '14 at 09:19

5

Whatever is the endianness of your system, your SerializeInt function will store value in big endian in the array buffer. It is stored as big endian as it writes the most significant byte first. Remember that when you evaluate value it has no endianness per se, it's like any mathematical value.

answered Mar 14 '14 at 09:19

ouah

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Yes my judgement was going in that direction too... Bit more elaboration would be welcome though as to why – Mar 14 '14 at 09:22
Yes, I was also thinking that it had to do with `value` ... like integer 14 is 14 both on BE and LE machines, ... that is the point right? (and doing `value >> 24` will always yield MSB?) – Mar 14 '14 at 09:49
@dmcr_code a value has no endianness, endianness is the way you'll store a value in a multi-bytes array. If `value` is a 32-bit quantity then yes `value >> 24` is the MSB whatever the endianness of the system is. – ouah Mar 14 '14 at 09:53

0xF1 · Answer 2 · 2014-03-14T09:26:57.360

The code will always write the integer in char array in big-endian way. This is because you are always saving MSB at buffer[0] and LSB at buffer[3].

Suppose in a little endian machine:

you have number 0x11223344: (in hexadecimal format)

Address      ->      2003  2002  2001  2000

`int` Number ->       11    22    33    44
Buffer       ->   index[3]  [2]   [1]   [0]

The same representation on a big endian machine will be:

Address      ->      2000  2001  2002  2003  <= See only the addressing changed/reversed

`int` Number ->       11    22    33    44
Buffer       ->   index[3]  [2]   [1]   [0]

In both machines, integer << 24 will mean MSB and (int8_t)integer will mean LSB.

score 0 · Answer 3 · answered Mar 14 '14 at 09:22

0

Big-Endian is defined as with the big end (i.e. most significant bit/byte) first. So, yes, the function writes the value in big-endian format to the buffer.

However, returning a pointer one past the buffer strikes me as unusual.

answered Mar 14 '14 at 09:22

Jens

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You can write the next value in the location immediately following what you just wrote – David Heffernan Mar 14 '14 at 09:25

score 0 · Answer 4 · answered Mar 14 '14 at 09:40

Apart from the already posted answers regarding endianess...

Does this function work correctly on both Little endian and Big endian machines?

No, because you use bit-wise arithmetic on the default int type. If the int is negative, the behavior is implementation-defined. In general, it doesn't make any sense whatsoever to use bit-wise operators on signed integers.

Furthermore, your code relies on int being 32 bits, of which there are no guarantees.

So your code is completely non-portable. Fix it in the following way:

#include <stdint.h>

uint8_t* SerializeInt (uint8_t* buffer, uint32_t value)

convert int to char array Big Endian

4 Answers4

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