How to add or multiply two i16 in Rust to form an i32 without overflowing?

Question

How to add or multiply two i16 numbers in Rust into a larger i32 without overflowing?

let a: i16 = i16::max_value();
let b: i16 = i16::max_value();
let c: i32 = a + b; // throws EO308 "Expected `u32`, found `u16`

Answer 1

There are no integral promotions or implicit casts in Rust, so you must do all type conversions manually.

For casts, you can use <value> as <type>. However, if the destination type is a superset of the original one and converting it cannot lose information (such as your case), you can self-document that by using <type>::from:

let a: i16 = i16::max_value();
let b: i16 = i16::max_value();
let c: i32 = i32::from(a) + i32::from(b);
assert_eq!(c, 65534);

This particular case cannot overflow, but for other cases you can prevent overflows using {integer}::checked_*() functions:

let a: i16 = i16::max_value();
let b: i16 = i16::max_value();
let c: Option<i16> = a.checked_add(b);
assert_eq!(c, None); //overflow!

Note that overflowing an integer operation panics by default in debug builds.

If what you want is wrapping arithmetic, just like in old C, you can use {integer}::wraping_*(), with the additional bonus that it works for signed as well as unsigned values.

let a: i16 = i16::max_value();
let b: i16 = i16::max_value();
let c: i16 = a.wrapping_add(b);
assert_eq!(c, -2);