How do I convert a big-endian struct to a little endian-struct?

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:

Answer 1

You also have to consider alignment differences between the two compilers. Each compiler is allowed to insert padding between members in a structure the best suits the architecture. So you really need to know:

• How the UNIX prog writes to the file
• If it is a binary copy of the object the exact layout of the structure.
• If it is a binary copy what the endian-ness of the source architecture.

This is why most programs (That I have seen (that need to be platform neutral)) serialize the data as a text stream that can be easily read by the standard iostreams.

Answer 2

Something like this should work:

#include <algorithm>

struct RECORD {
    UINT32 foo;
    UINT32 bar;
    CHAR fooword[11];
    CHAR barword[11];
    UINT16 baz;
}

void ReverseBytes( void *start, int size )
{
    char *beg = start;
    char *end = beg + size;

    std::reverse( beg, end );
}

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

    // for each entry {
    RECORD r;
    f.read( (char *)&r, sizeof( RECORD ) );
    ReverseBytes( r.foo, sizeof( UINT32 ) );
    ReverseBytes( r.bar, sizeof( UINT32 ) );
    ReverseBytes( r.baz, sizeof( UINT16 )
    // }

    return 0;
}