[ACCEPTED]-Reading a C/C++ data structure in C# from a byte array-marshalling
From what I can see in that context, you 5 don't need to copy SomeByteArray into a buffer. You simply 4 need to get the handle from SomeByteArray, pin it, copy 3 the IntPtr data using PtrToStructure and then release. No need 2 for a copy.
That would be:
NewStuff ByteArrayToNewStuff(byte[] bytes)
{
GCHandle handle = GCHandle.Alloc(bytes, GCHandleType.Pinned);
try
{
NewStuff stuff = (NewStuff)Marshal.PtrToStructure(handle.AddrOfPinnedObject(), typeof(NewStuff));
}
finally
{
handle.Free();
}
return stuff;
}
Generic version:
T ByteArrayToStructure<T>(byte[] bytes) where T: struct
{
T stuff;
GCHandle handle = GCHandle.Alloc(bytes, GCHandleType.Pinned);
try
{
stuff = (T)Marshal.PtrToStructure(handle.AddrOfPinnedObject(), typeof(T));
}
finally
{
handle.Free();
}
return stuff;
}
Simpler 1 version (requires unsafe switch):
unsafe T ByteArrayToStructure<T>(byte[] bytes) where T : struct
{
fixed (byte* ptr = &bytes[0])
{
return (T)Marshal.PtrToStructure((IntPtr)ptr, typeof(T));
}
}
Here is an exception safe version of the 1 accepted answer:
public static T ByteArrayToStructure<T>(byte[] bytes) where T : struct
{
var handle = GCHandle.Alloc(bytes, GCHandleType.Pinned);
try {
return (T) Marshal.PtrToStructure(handle.AddrOfPinnedObject(), typeof(T));
}
finally {
handle.Free();
}
}
Watch out for packing issues. In the example 4 you gave all fields are at the obvious offsets 3 because everything is on 4 byte boundaries 2 but this will not always be the case. Visual 1 C++ packs on 8 byte boundaries by default.
object ByteArrayToStructure(byte[] bytearray, object structureObj, int position)
{
int length = Marshal.SizeOf(structureObj);
IntPtr ptr = Marshal.AllocHGlobal(length);
Marshal.Copy(bytearray, 0, ptr, length);
structureObj = Marshal.PtrToStructure(Marshal.UnsafeAddrOfPinnedArrayElement(bytearray, position), structureObj.GetType());
Marshal.FreeHGlobal(ptr);
return structureObj;
}
Have this
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