C++ Reading bytes Problem

Question

I have this issue, I don't know if is expected, here is the thing:

I'm trying to load bytes from a file to an structure like this:

struct
{
    char
    char
    char
    char
    unsigned int
}

but the problem is when the unsigned int is filled, it seems like in the reading stream the bytes are swapped, e.x. if the file contains 0x45080000, the unsigned int will have 0x84500000, which is wrong.

This can be "solved" if i change the unsigned int for a BYTE[4], but is not what I want. Here is the code that I use to read from the file:

fopen_s( &mFile, "myFile.ext", "rb" );

if( mFile == NULL ) print( " **** E R R O R ! **** " );
else
{
    if( fread( &myStruct, sizeof( MY_Struct ), 1, myFile) != 1)
    {
        print( " **** E R R O R ! **** " );
        return 0;
    }
}

Is this an expected behavior or what am I doing wrong?

Regards

I dont know if this is a problem of endianness because I'm just reading the bytes as they are in the file stored, one by one, so endianness should not affect, I think — cbedoya
– cbedoya, Commented Jul 26, 2011 at 19:36
lol, i just put error because is an error :P, to be a little generic xD — cbedoya
– cbedoya, Commented Jul 26, 2011 at 19:37

Mark B · Accepted Answer · 2011-07-26 19:00:27Z

3

As you've discovered, portable serialization can be a pain. Instead of writing and reading the structure, you need to serialize each attribute individually in a normalized format (network byte order is common). Then when you deserialize the bytes come back correctly.

answered Jul 26, 2011 at 19:00

Mark B

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1 Comment

cbedoya Over a year ago

ups, then i think i will prepare for all the pain =(

Tom Kerr · Accepted Answer · 2011-07-26 19:10:42Z

1

Either serialize/unserialize each field independently (standards conforming) or...

use a platform specific option:

#pragma pack(push,1)
struct foo {
  // ...
};
#pragma pack(pop)

This aligns all variables in foo to 1 byte alignment, so it wont work for a boolean.

If you intend to be cross-platform you'll have to test the hell out of it for problems of endianess and pragma support.

edited Jul 26, 2011 at 19:10

answered Jul 26, 2011 at 18:59

Tom Kerr

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5 Comments

cbedoya Over a year ago

yeah, it will be cross-platform, but i cannot serialize each field, because if I have a big file, and i need to grab some info of it, or you think the best solution is to read byte by byte?

Tom Kerr Over a year ago

@user864094 It's hard to answer that without knowing about structure of the file. You probably need to describe what you are trying to do more.

cbedoya Over a year ago

is a file, that stores some info about frames and modules, like a sprite editor, this file have a header with the first 4 bytes indicating a name, one more byte indicating if is a module or animation or frame, and then, the unsigned int that is a size to reading info, but, the problem comes here, because as I said, if the size is 0x45080000, the program itself put the 08 at the beginning, but this not happens when I read byte by byte, only when I change the types

Tom Kerr Over a year ago

Reading a binary blob into a struct is not standards compliant. There may be some endianess problems as the others have mentioned. If byte by byte is working, then I would go with it. If the subtleties of our solutions don't make any sense then it probably requires more research.

cbedoya Over a year ago

ok, at the moment I will go for the byte by byte reading, as you say, this will need more research, thanx for your answer

minus · Accepted Answer · 2011-07-27 14:38:34Z

1

Just to add a bit of icing to the cake!

You must handle 32 vs. 64 bit issues also.

A very good include file is stdint.h (C99). It defines types with specific size so you get less problems when switching the word width.

answered Jul 27, 2011 at 14:38

minus

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Comments

oliver · Accepted Answer · 2011-07-26 19:06:22Z

0

you probably need to take into account big/little endian. try s.th. like this:

#define BIG_ENDIAN
//#define LITTLE_ENDIAN
...
#ifdef BIG_ENDIAN

inline void writeMem32(uint8* buf, uint32 val)
{
    buf[0] = static_cast<uint8>(val >> 24);
    buf[1] = static_cast<uint8>(val >> 16);
    buf[2] = static_cast<uint8>(val >>  8);
    buf[3] = static_cast<uint8>(val >>  0);
}

inline uint32 readMem32(const uint8* buf)
{
    return ( ((*(buf+0)) << 24) | ((*(buf+1)) << 16) | ((*(buf+2)) << 8) | (*(buf+3)) );
}

#endif // BIG_ENDIAN

if you are on little endian the byte order is of course different :)

answered Jul 26, 2011 at 19:06

oliver

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4 Comments

cbedoya Over a year ago

may be this is the problem, if the file is written in little endian, the bytes are not readed in the same order?

RocketR Over a year ago

There's standard hton/ntoh functions. Don't reinvent the wheel.

oliver Over a year ago

@RocketR: I agree. But I found that sometimes it's useful to reinvent just to get a deeper understanding of what is really going on...that's at least what is true for myself :)

oliver Over a year ago

@user864094: depends on the architecture of the system where you read the file (this has some examples). In your comment you say first 4 bytes indicating a name...one more byte indicating if is a module or animation or frame, and then, the unsigned int that is a size But your struct only contains 4 bytes and 1 integer...maybe you are missing a byte here. Also: The layout of your struct is absolutely platform-dependent. You have to check how your compiler does the alignment (might turn out that each bytes gets aligned at 32bit boundaries).

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