Why do C#'s binary operators always return int regardless of the format of their inputs?

Question

If I have two bytes a and b, how come:

byte c = a & b;

produces a compiler error about casting byte to int? It does this even if I put an explicit cast in front of a and b.

Also, I know about this question, but I don't really know how it applies here. This seems like it's a question of the return type of operator &(byte operand, byte operand2), which the compiler should be able to sort out just like any other operator.

Answer 1

Why do C#'s bitwise operators always return int regardless of the format of their inputs?

I disagree with always. This works and the result of a & b is of type long:

long a = 0xffffffffffff;
long b = 0xffffffffffff;
long x = a & b;

The return type is not int if one or both of the arguments are long, ulong or uint.

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Why do C#'s bitwise operators return int if their inputs are bytes?

The result of byte & byte is an int because there is no & operator defined on byte. (Source)

An & operator exists for int and there is also an implicit cast from byte to int so when you write byte1 & byte2 this is effectively the same as writing ((int)byte1) & ((int)byte2) and the result of this is an int.