byte + byte = int… why?

Question

Looking at this C# code:

byte x = 1;
byte y = 2;
byte z = x + y; // ERROR: Cannot implicitly convert type 'int' to 'byte'

The result of any math performed on byte (or short) types is implicitly cast back to an integer. The solution is to explicitly cast the result back to a byte:

byte z = (byte)(x + y); // this works

What I am wondering is why? Is it architectural? Philosophical?

We have:

• int + int = int
• long + long = long
• float + float = float
• double + double = double

So why not:

• byte + byte = byte
• short + short = short?

A bit of background: I am performing a long list of calculations on "small numbers" (i.e. < 8) and storing the intermediate results in a large array. Using a byte array (instead of an int array) is faster (because of cache hits). But the extensive byte-casts spread through the code make it that much more unreadable.

Answer 1

This is for the most part my answer that pertains to this topic, submitted first to a similar question here.

All operations with integral numbers smaller than Int32 are rounded up 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".

If you desire the result back in Int16 format, it is irrelevant if you perform the cast in code, or the compiler (hypotetically) emits the conversion "under the hood".

For example, to do Int16 arithmetic:

short a = 2, b = 3;

short c = (short) (a + b);

The two numbers would expand to 32 bits, get added, then truncated back to 16 bits, which is how MS intended it to be.

The advantage of using short (or byte) is primarily storage in cases where you have massive amounts of data (graphical data, streaming, etc.)

Answer 2

Addition is not defined for bytes. So they are cast to int for the addition. This true for most math operations and bytes. (note this is how it used to be in older languages, I am assuming that it hold true today).

Answer 3

In addition to all the other great comments, I thought I would add one little tidbit. A lot of comments have wondered why int, long, and pretty much any other numeric type doesn't also follow this rule...return a "bigger" type in response to arithmatic.

A lot of answers have had to do with performance (well, 32bits is faster than 8bits). In reality, an 8bit number is still a 32bit number to a 32bit CPU....even if you add two bytes, the chunk of data the cpu operates on is going to be 32bits regardless...so adding ints is not going to be any "faster" than adding two bytes...its all the same to the cpu. NOW, adding two ints WILL be faster than adding two longs on a 32bit processor, because adding two longs requires more microops since you're working with numbers wider than the processors word.

I think the fundamental reason for causing byte arithmetic to result in ints is pretty clear and straight forward: 8bits just doesn't go very far! :D With 8 bits, you have an unsigned range of 0-255. That's not a whole lot of room to work with...the likelyhood that you are going to run into a bytes limitations is VERY high when using them in arithmetic. However, the chance that you're going to run out of bits when working with ints, or longs, or doubles, etc. is significantly lower...low enough that we very rarely encounter the need for more.

Automatic conversion from byte to int is logical because the scale of a byte is so small. Automatic conversion from int to long, float to double, etc. is not logical because those numbers have significant scale.

Answer 4

I thought I had seen this somewhere before. From this article, The Old New Thing:

Suppose we lived in a fantasy world where operations on 'byte' resulted in 'byte'.

byte b = 32;
byte c = 240;
int i = b + c; // what is i?

In this fantasy world, the value of i would be 16! Why? Because the two operands to the + operator are both bytes, so the sum "b+c" is computed as a byte, which results in 16 due to integer overflow. (And, as I noted earlier, integer overflow is the new security attack vector.)

EDIT: Raymond is defending, essentially, the approach C and C++ took originally. In the comments, he defends the fact that C# takes the same approach, on the grounds of language backward compatibility.