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

Solution 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.

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.

Solution 2:

This behavior is a consequence of the design of IL, the intermediate language generated by all .NET compilers. While it supports the short integer types (byte, sbyte, short, ushort), it has only a very limited number of operations on them. Load, store, convert, create array, that's all. This is not an accident, those are the kind of operations you could execute efficiently on a 32-bit processor, back when IL was designed and RISC was the future.

The binary comparison and branch operations only work on int32, int64, native int, native floating point, object and managed reference. These operands are 32-bits or 64-bits on any current CPU core, ensuring the JIT compiler can generate efficient machine code.

You can read more about it in the Ecma 335, Partition I, chapter 12.1 and Partition III, chapter 1.5

I wrote a more extensive post about this over here.

Solution 3:

Binary operators are not defined for byte types (among others). In fact, all binary (numeric) operators act only on the following native types:

• int
• uint
• long
• ulong
• float
• double
• decimal

If there are any other types involved, it will use one of the above.

It's all in the C# specs version 5.0 (Section 7.3.6.2):

Binary numeric promotion occurs for the operands of the predefined +, –, *, /, %, &, |, ^, ==, !=, >, <, >=, and <= binary operators. Binary numeric promotion implicitly converts both operands to a common type which, in case of the non-relational operators, also becomes the result type of the operation. Binary numeric promotion consists of applying the following rules, in the order they appear here:

• If either operand is of type decimal, the other operand is converted to type decimal, or a compile-time error occurs if the other operand is of type float or double.
• Otherwise, if either operand is of type double, the other operand is converted to type double.
• Otherwise, if either operand is of type float, the other operand is converted to type float.
• Otherwise, if either operand is of type ulong, the other operand is converted to type ulong, or a compile-time error occurs if the other operand is of type sbyte, short, int, or long.
• Otherwise, if either operand is of type long, the other operand is converted to type long.
• Otherwise, if either operand is of type uint and the other operand is of type sbyte, short, or int, both operands are converted to type long.
• Otherwise, if either operand is of type uint, the other operand is converted to type uint.
• Otherwise, both operands are converted to type int.