How to copy array to structure with fixed array?

Question

What is the best way for Copying 'mydata' to 'sample_a'?

I expect to get 4 in sample_a.data_array[ 1 ]. That value is p[ 9 ].

I plan to use this for TCP/UDP packet communication.

[StructLayout(LayoutKind.Sequential, Pack = 1)]
struct SampleA
{
    public int data1;
    public int data2;
    public unsafe fixed char data_array[3];
    public int data3;
}

static void Main(string[] args)
{
    Console.WriteLine("Hello World!");

    short bytesize = 15; 
    byte[] mydata = new byte[bytesize];
    unsafe
    {
        fixed (byte* p = mydata)
        {
            // data1
            p[0] = 1;
            // data2
            p[4] = 2;
            // data_array
            p[8] = 3;
            p[9] = 4;
            p[10] = 5;
            // data3
            p[11] = 6;
        }

        // Copy 'mydata' to 'sample_a'
        IntPtr intptr = Marshal.AllocHGlobal(bytesize);
        Marshal.Copy(mydata, 0, intptr, bytesize);
        SampleA sample_a = (SampleA)Marshal.PtrToStructure(intptr, typeof(SampleA));
        Marshal.FreeHGlobal(intptr);

        // I want to get 4 in a1
        int a1 = (int)sample_a.data_array[1];
    }
}

You could also check the MemoryMarshal.Cast this doesn't do a memory copy. — Jeroen van Langen
– Jeroen van Langen, Commented Feb 5, 2020 at 15:13

Marc Gravell · Accepted Answer · 2020-02-05 15:26:35Z

3

You can avoid a lot of work here with spans:


// note: size is 18; if you intended 15 - perhaps change char=>byte
byte[] mydata = new byte[Unsafe.SizeOf<SampleA>()];
// alternative: Span<byte> mydata = stackalloc byte[Unsafe.SizeOf<SampleA>()];

// ..TODO: populate mydata

// cast it out
var sample_a = MemoryMarshal.Cast<byte, SampleA>(mydata)[0];

You can also use Unsafe.As, but this is IMO not a good idea as it won't detect incorrect memory conditions as well:

var sample_a = Unsafe.As<byte, SampleA>(ref mydata[0]);

edited Feb 5, 2020 at 15:26

answered Feb 5, 2020 at 15:16

Marc Gravell

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4 Comments

canton7 Over a year ago

Note also that this doesn't use the marshaller, and so none of the attributes which normally affect the marshaller, e.g. MarshalAs, will do anything.

Marc Gravell Over a year ago

@canton7 not true; StructLayout also impacts the layout of the struct itself: so yes, it still has effect (it won't impact MarshalAs - that was a comment edit, though!)

canton7 Over a year ago

True, bad example on my side. Changed.

Marc Gravell Over a year ago

@canton7 re the bytes vs char: yes, that is true; if OP intends these to be bytes: they should declare them as bytes

canton7 · Accepted Answer · 2020-02-05 15:27:55Z

2

Your problem is that C#'s chars are 2 bytes wide, and you're treating them as if they're 1 byte wide.

One fix is to explicitly declare them as byte:

[StructLayout(LayoutKind.Sequential, Pack = 1)]
struct SampleA
{
    public int data1;
    public int data2;
    public unsafe fixed byte data_array[3];
    public int data3;
}

It's more common to embed arrays using UnmanagedType.ByValArray:

[StructLayout(LayoutKind.Sequential, Pack = 1)]
struct SampleA
{
    public int data1;
    public int data2;
    [MarshalAs(UnmanagedType.ByValArray, SizeConst = 3)]
    public byte[] data_array;
    public int data3;
}

If you want to use char, you can tell the marshaller that each should be marshalled as a U8:

[StructLayout(LayoutKind.Sequential, Pack = 1)]
struct SampleA
{
    public int data1;
    public int data2;
    [MarshalAs(UnmanagedType.ByValArray, SizeConst = 3, ArraySubType = UnmanagedType.U8)]
    public char[] data_array;
    public int data3;
}

You can also tell the marshaller that you're using ANSI strings, which has the same effect

[StructLayout(LayoutKind.Sequential, Pack = 1, CharSet = CharSet.Ansi)]
struct SampleA
{
    public int data1;
    public int data2;
    [MarshalAs(UnmanagedType.ByValArray, SizeConst = 3)]
    public char[] data_array;
    public int data3;
}

edited Feb 5, 2020 at 15:27

answered Feb 5, 2020 at 15:01

canton7

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1 Comment

Marc Gravell Over a year ago

That is one of several ways of doing this; however, it is not the only option

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