Question

C++ code is :

unsigned short* ui=(unsigned short*)&buf[110];
            CountDev=ui[0];

buf is byte[] and CountDev is unsigned int

(BCB6 Compiler x86)

My try is :
F#

...CountDev  = System.BitConverter.ToInt32( [| arrayRead.[110]; arrayRead.[111] |] , 0 )

C#

...CountDev  = System.BitConverter.ToInt32( [arrayRead[110]; arrayRead[111]] , 0 )

But seriously I can’t be sure about it. Check my try and tell me if I am doing it wrong please.

Answer 1

I would just do this to simply concatenate the two bytes and putting it into an int:

UInt32 CountDev = (UInt32)arrayRead[111] << 8 | (UInt32)arrayRead[110];

since you just need the least significant two byte, and int is 4 byte long (the most significant or sign bit is not touched), you can also use a signed int:

int CountDev = (int)arrayRead[111] << 8 | (int)arrayRead[110];

Edit:

Henk Holtermans solution is definitely the better choice as it uses the endianess of the current machine:

UInt32 CountDev = (UInt32)System.BitConverter.ToUint16(arrayRead, 110);