Why are there no arithmetic operators (+,-,*,/,%) for 8 and 16 bit integers in C#? [duplicate]

Question

I got shocked knowing that there are no arithmetic operator +, -, *, /, and % for 8 and 16 bit integers in C#. I am reading "C# 5.0 Pocket Reference" on page 23 as follows.

The following code does not compile.

class Program
{
    static void With16Bit()
    {
        short a = 1;
        short b = 2;
        short c = a + b;
        Console.WriteLine(c);
    }

    static void With8Bit()
    {
        byte a = 1;
        byte b = 2;
        byte c = a + b;
        Console.WriteLine(c);
    }

    static void Main(string[] args)
    {
        With8Bit();
        With16Bit();
    }
}

Why did the C# designers do it? What are their consideration about it?

Answer 1

There are arithmetics with Int8, Int16; but the result is int32 and so you have to cast:

class Program
{
    static void With16Bit()
    {
        short a = 1;
        short b = 2;
        short c = (short) (a + b); // <- cast, since short + short = int
        Console.WriteLine(c);
    }

    static void With8Bit()
    {
        byte a = 1;
        byte b = 2;
        byte c = (byte) (a + b); // <- cast, since byte + byte = int
        Console.WriteLine(c);
    }

    static void Main(string[] args)
    {
        With8Bit();
        With16Bit();
    }
}

Answer 2

Plese remember that when you perform addition operation on short and byte the default outcome would be Integer.

So the :

byte + byte = int
short + short = int

So if you want to get back the actual value you need to cast it back.

Try This:

       short a = 1;
       short b = 2;
       short c =(short) (a + b);
       Console.WriteLine(c);

       byte a = 1;
       byte b = 2;
       byte c =(byte) (a + b);
       Console.WriteLine(c);

From The Source:

This Behaviour is because the designers have not considered that byte and short as actual numbers ,but they have considered them as only sequence of bits. so performing arithmetic operations on them does not make any sense so if that is the case int and long would serve the purpose.

Answer 3

From kek444 answer

All operations with integral numbers smaller than Int32 are widened to 32 bits 
before calculation by default. The reason why the result is Int32 is
simply to leave it as it is after calculation. If you check the
MSIL arithmetic opcodes, the only integral numeric type they operate 
with are Int32 and Int64. It's "by design".