C

retnuoc · 07-03-2008, 09:05 PM

Hello, i meet a C problem recently, a union:

union u_tag {
char c;
int i;
}tag;

let's say i store a value into tag.i (int), after that i retrieve the value by call tag.c (char), is tag.c (char) taking the LSB/MSB value of tag.i (int)? else machine and compiler determine the result?

estabroo · 07-03-2008, 09:50 PM

That'll be machine dependent because endianess will switch which byte its looking at. If you look at it like its flat memory with the integer bytes starting at 0 and the character byte being at 0 then in little endian it'll be LSB and in big endian it'll be MSB that gets mapped to location 0.

For example if the integer was assigned 0x04030201

on a little endian
byte position
0 1 2 3
01 02 03 04

on a big endian
byte position
0 1 2 3
04 03 02 01

Wim Sturkenboom · 07-04-2008, 12:51 AM

As stated by estabroo, it's machine dependent.

Your solution will either be in conditional compiling (ifdefs) based on the machine (i386, 68000, ...) or you can have a look at the functions that convert integers from host byte order to network byte order (htonl/htons and the reverse ntohl/ntohs).

WimS

Wim Sturkenboom · 07-04-2008, 02:12 PM

On another note:
I thought that the elements of an union we're supposed to be of the same size (so char[2] or char[4] in your case); I might however be wrong

dmail · 07-04-2008, 03:15 PM

Quote:

Originally Posted by Wim Sturkenboom

On another note:
I thought that the elements of an union we're supposed to be of the same size (so char[2] or char[4] in your case); I might however be wrong

I do not understand what you mean by the highlighted text, but the size of elements in a union does not matter the union will use a type which can hold the largest element IIRC.

Mr. C. · 07-04-2008, 03:26 PM

Yes, a union will occupy the largest space to accommodate the largest member.

Wim Sturkenboom · 07-05-2008, 03:39 AM

Sorry, I've always assumed that if you have 16 (8-bit) chars in a union, you also needed to have 4 32-bit integers in there.

From your comments, it's obvious that I was wrong.

Mr. C. · 07-05-2008, 12:25 PM

Since structure alignment can affect an applications performance, sometimes it is best to ensure structures are properly aligned the hardwares natural size boundaries. Padding a union is sometimes a good idea in such cases. But the language ensures that the types in the union are accessible regardless of union member size differences.

retnuoc · 07-07-2008, 01:51 AM

Quote:

Originally Posted by estabroo

That'll be machine dependent because endianess will switch which byte its looking at. If you look at it like its flat memory with the integer bytes starting at 0 and the character byte being at 0 then in little endian it'll be LSB and in big endian it'll be MSB that gets mapped to location 0.

For example if the integer was assigned 0x04030201

on a little endian
byte position
0 1 2 3
01 02 03 04

on a big endian
byte position
0 1 2 3
04 03 02 01

Thanks estabroo, i got a clearer picture about union d.