CODEWITHSUNDEEP
c++binaryfilesendiannessreinterpret-cast
Vincent
Vincent

Reputation: 60341

Binary read, reinterpret_cast and endianness

I'm currently dealing with endianness-related problems.

Let's assume that I have a big-endian file in a big-endian system.

The first value in this file is 2882400152 = 0xABCDEF98 which is an integer 4.

To read that value in an integer 4, _myint4, I simply do : 

mystream.read(reinterpret_cast<char*>(&_myint4), sizeof(_myint4))

The question is : what is the equivalent to read the integer 4 value in the file, in an integer 8, _myint8 ?

My first guess would be something like that :

mystream.read((reinterpret_cast<char*>(&_myint8))+4, 4); // big-endian file/system
mystream.read(reinterpret_cast<char*>(&_myint8), 4); // little-endian file/system

But I'm not sure at all. What is the good way to do this ?

IMPORTANT : I cannot use a temporary integer 4 value, I need to read directly the integer 4 in _myint8.

Upvotes: 1

Views: 3476

Answers (1)

DS.
DS.

Reputation: 24110

Your guess seems right:

_myint8 = 0;
mystream.read((reinterpret_cast<char*>(&_myint8))+4, 4); // big-endian file/system
mystream.read(reinterpret_cast<char*>(&_myint8), 4); // little-endian file/system

To figure out byte ordering on your own, it might help to play a bit in Python:

>>> import struct
>>> struct.pack(">Q", 0xabcdef98)    // Big-endian 8-byte int.
'\x00\x00\x00\x00\xab\xcd\xef\x98'   // Its layout in memory.
>>> struct.pack(">I", 0xabcdef98)    // Big-endian 4-byte int.
'\xab\xcd\xef\x98'

>>> struct.pack("<Q", 0xabcdef98)    // Little-endian 8-byte int.
'\x98\xef\xcd\xab\x00\x00\x00\x00'
>>> struct.pack("<I", 0xabcdef98)    // Little-endian 4-byte int.
'\x98\xef\xcd\xab'

So you are right, you just need to ensure you have zeroes in the places in memory that you aren't overriding.

Upvotes: 2

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