Question

Could someone point me, why here:

Byte b = 100; 
b = (Byte)(b+200);

I have to use explicit type conversion. But here

Byte b = 100;
b += 200;

I don’t need to do this?

Does compiler generate different IL code for this two cases? And which case is better?

Answer 1

Because the standard permits it (see the second case below):

14.14.2 Compound assignment

An operation of the form x op= y is processed by applying binary operator overload resolution (§14.2.4) as if the operation was written x op y. Then,

• If the return type of the selected operator is implicitly convertible to the type of x, the operation is evaluated as x = x op y, except that x is evaluated only once.

• Otherwise, if the selected operator is a predefined operator, if the return type of the selected operator is explicitly convertible to the type of x, and if y is implicitly convertible to the type of x or the operator is a shift operator, then the operation is evaluated as x = (T)(x op y), where T is the type of x, except that x is evaluated only once.

• Otherwise, the compound assignment is invalid, and a compile-time error occurs.

The IL code should be essentially identical in this case. Of course, if evaluating b has side effects, it will be evaluated twice in the b = (byte)b + 200 case and only once when using compound assignment.