Question

I have a binary file that was created on a unix machine. It’s just a bunch of records written one after another. The record is defined something like this:

struct RECORD {
  UINT32 foo;
  UINT32 bar;
  CHAR fooword[11];
  CHAR barword[11];
  UNIT16 baz;
}

I am trying to figure out how I would read and interpret this data on a Windows machine. I have something like this:

fstream f;
f.open("file.bin", ios::in | ios::binary);

RECORD r;

f.read((char*)&detail, sizeof(RECORD));

cout << "fooword = " << r.fooword << endl;

I get a bunch of data, but it’s not the data I expect. I’m suspect that my problem has to do with the endian difference of the machines, so I’ve come to ask about that.

I understand that multiple bytes will be stored in little-endian on windows and big-endian in a unix environment, and I get that. For two bytes, 0x1234 on windows will be 0x3412 on a unix system.

Does endianness affect the byte order of the struct as a whole, or of each individual member of the struct? What approaches would I take to convert a struct created on a unix system to one that has the same data on a windows system? Any links that are more in depth than the byte order of a couple bytes would be great, too!

Answer 1

As well as the endian, you need to be aware of padding differences between the two platforms. Particularly if you have odd length char arrays and 16 bit values, you may well find different numbers of pad bytes between some elements.

Edit: if the structure was written out with no packing, then it should be fairly straightforward. Something like this (untested) code should do the job:

// Functions to swap the endian of 16 and 32 bit values

inline void SwapEndian(UINT16 &val)
{
    val = (val<<8) | (val>>8);
}

inline void SwapEndian(UINT32 &val)
{
    val = (val<<24) | ((val<<8) & 0x00ff0000) |
          ((val>>8) & 0x0000ff00) | (val>>24);
}

Then, once you’ve loaded the struct, just swap each element:

SwapEndian(r.foo);
SwapEndian(r.bar);
SwapEndian(r.baz);