Assigning an int to a struct object without using unsafe

Question

I have a struct definition in c# as follows

public struct test { byte SetCommonPOP; byte SetCommonSVP; byte SetCommonUHDP; byte SetCommonMHDP; };

How do I assign an int y to an object x of this structure without having to use unsafe?

How would you want the int to be interpreted? Different endianness systems could convert in different ways. — Jon Skeet
– Jon Skeet, Commented Oct 21, 2013 at 16:41
Both an int and the struct object takes 4 bytes. Assuming that the first member takes up the LSB and the fourth the MSB — Tyler Durden
– Tyler Durden, Commented Oct 21, 2013 at 16:42
That would be big Endian right? Sorry, but I always get confused. — Tyler Durden
– Tyler Durden, Commented Oct 21, 2013 at 16:50
I believe that's little-endian, actually: en.wikipedia.org/wiki/Endianness Anyway, that's why my answer makes it explicit - you can always change it round :) — Jon Skeet
– Jon Skeet, Commented Oct 21, 2013 at 16:53
That would be little-endian. I remember it by the story that is the source of the "endian" words: those who start with the little end of the soft-boiled eggs are little-endian, those who start with the big end are big-endian. It's confusing because big-endian isn't when you end with the big values, but when you start with the "big end". — Tim S.
– Tim S., Commented Oct 21, 2013 at 16:53

Jon Skeet · Accepted Answer · 2013-10-21 16:45:15Z

You could write a custom conversion operator:

public struct Test { private readonly byte pop; private readonly byte svp; private readonly byte uhdp; private readonly byte mhdp; // TODO: Properties to return the above public Test(byte pop, byte svp, byte uhdp, byte mhdp) { this.pop = pop; this.svp = svp; this.uhdp = uhdp; this.mhdp = mhdp; } public static implicit operator Test(int value) { // Working from most significant to least significant bits... byte mhdp = (byte) ((value >> 0) & 0xff); byte uhdp = (byte) ((value >> 8) & 0xff); byte svp = (byte) ((value >> 16) & 0xff); byte pop = (byte) ((value >> 24) & 0xff); return new Test(pop, svp, uhdp, mhdp); } }

Personally I'd prefer a static FromInt32 method instead of an implicit operator, but that's your call. It's very possible that you don't need all the & 0xff parts in the conversion - and I wouldn't bother with them if you were using a uint instead of an int. Extracting parts of a signed integer just makes me twitchy, and this is possibly overcompensation. Another alternative would be a cast of value to a uint as a local variable to start with.

Eli Arbel · Accepted Answer · 2013-10-21 17:12:26Z

Another option is to use an explicit struct layout:

[StructLayout(LayoutKind.Explicit)] public struct Test { [FieldOffset(3)] public readonly byte pop; [FieldOffset(2)] public readonly byte svp; [FieldOffset(1)] public readonly byte uhdp; [FieldOffset(0)] public readonly byte mhdp; [FieldOffset(0)] private int value; public static Test FromInt32(int value) { var test = new Test(); test.value = value; return test; } }

Ok, so here, it is treated like a union with value member occupying the same space as the first 4 members. Right?
Should work too I guess. Only thing I dont want to alter the struct definition. Thanks anyways

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