endianness

I need to hand assemble 14bit MIDI Pitch Bend values from raw UInt16 values in iOS. I'm wondering if anybody out there has had a chance to come up with an elegant solution? Here's where I'm at - I'll get a chance to test this probably later today, but if I hear back before then, great: First, some MIDI preliminaries for anybody curious. MIDI Pitch Bend is broken up into one Status Byte followed by two Data Bytes (it's a 14bit controller), these two Data Bytes are associated with their Status Byte by both leading with a Zero status bit , MIDI Spec has them appearing in the order of MSB -> LSB

Write a program to determine whether a computer is big-endian or little-endian. bool endianness() { int i = 1; char *ptr; ptr = (char*) &i; return (*ptr); } So I have the above function. I don't really get it. ptr = (char*) &i, which I think means a pointer to a character at address of where i is sitting, so if an int is 4 bytes, say ABCD, are we talking about A or D when you call char* on that? and why? Would some one please explain this in more detail? Thanks. So specifically, ptr = (char*) &i; when you cast it to char*, what part of &i do I get? If you have a little-endian architecture, i

Is there an option to find if my system is little endian byte order or big endian byte order using Perl? perl -MConfig -e 'print "$Config{byteorder}\n";' See Perl documentation . Sean I guess you could do: $big_endian = pack("L", 1) eq pack("N", 1); This might fail if your system has a nonstandard (neither big-endian nor little-endian) byte ordering (eg PDP-11). 来源： https://stackoverflow.com/questions/2610849/finding-if-the-system-is-little-endian-or-big-endian-with-perl

Im trying to cast a 4 byte array to an ulong in C#. I'm currently using this code: atomSize = BitConverter.ToUInt32(buffer, 0); The byte[4] contains this: 0 0 0 32 However, the bytes are Big-Endian. Is there a simple way to convert this Big-Endian ulong to a Little-Endian ulong? Mark Byers I believe that the EndianBitConverter in Jon Skeet's MiscUtil library ( nuget link ) can do what you want. You could also swap the bits using bit shift operations: uint swapEndianness(uint x) { return ((x & 0x000000ff) << 24) + // First byte ((x & 0x0000ff00) << 8) + // Second byte ((x & 0x00ff0000) >> 8) +

I'm currently working on a big project that involve pictures. One of the big issues I'm having is with the endianness of the picture (jpeg to be clearer). I always though that in our modern world we didn't have to botter about this subject, but now I'm not sure. What I do: I do an HTTP request to an IP camera, the camera return me an array of byte. I parse these byte into an object Image in .NET using Image.FromStream. I take my Image object and do a Save to a physical file on the hard disk. These picture are then use in another module from my application which use a 3rd party "viewer" that do

union test { int i; char ch; }t; int main() { t.ch=20; } Suppose sizeof(int)==2 and let the memory addresses allocated for t are 2000, 2001. Then where is 20 i.e. t.ch stored - at 2000 or 2001 or depends on endianness of machine? The C99 standard (§6.7.2.1.14) says: The size of a union is sufficient to contain the largest of its members. The value of at most one of the members can be stored in a union object at any time. A pointer to a union object, suitably converted, points to each of its members (or if a member is a bit- field, then to the unit in which it resides), and vice versa.

Take this example: i = 0x12345678 print("{:08x}".format(i)) # shows 12345678 i = swap32(i) print("{:08x}".format(i)) # should print 78563412 What would be the swap32-function() ? Is there a way to byte-swap an int in python, ideally with built-in tools? One method is to use the struct module: def swap32(i): return struct.unpack("<I", struct.pack(">I", i))[0] First you pack your integer into a binary format using one endianness, then you unpack it using the other (it doesn't even matter which combination you use, since all you want to do is swap endianness). Big endian means the layout of a 32