Question

If I do the following in code for an 8-bit processor:

typedef union
{
unsigned long longn ;
unsigned char chars[4];
} longbytes;

Is longbytes.chars[0] always going to be the lowest byte of longbytes.longn, or does it depend on endianness/compiler/platform/target/luck etc.? I’ve viewed the disassembly of my complied code and that’s how it is in my specific case, but I’m curious if this code is portable.

Answer 1

There are several reasons why this is not portable:

• It depends on the endianess your platform (or compiler) enforces which byte is written first, so you can’t count on chars[0] addressing the lowest byte
• unsigned long is not guaranteed to be exactly as long as 4 chars, so depending on the platform you might not even get the complete long (or sizeof(long) might be smaller then 4 and you read further, but that’s unlikely for 8Bit processors at least.
• Reading a different union member then you wrote to is generally not portable, it is implementation defined behaviour. The reason for this is basically the combination of the two other issues.

So all in all that code is not portable at all.