Equality

Move supports two equality operations == and !=

Operations

Typing

Both the equal (==) and not-equal (!=) operations only work if both operands are the same type

0 == 0; // `true`
1u128 == 2u128; // `false`
b"hello" != x"00"; // `true`

Equality and non-equality also work over user defined types!

address 0x42 {
module example {
    struct S has copy, drop { f: u64, s: vector<u8> }

    fun always_true(): bool {
        let s = S { f: 0, s: b"" };
        // parens are not needed but added for clarity in this example
        (copy s) == s
    }

    fun always_false(): bool {
        let s = S { f: 0, s: b"" };
        // parens are not needed but added for clarity in this example
        (copy s) != s
    }
}
}

If the operands have different types, there is a type checking error

1u8 == 1u128; // ERROR!
//     ^^^^^ expected an argument of type 'u8'
b"" != 0; // ERROR!
//     ^ expected an argument of type 'vector<u8>'

Typing with references

When comparing references, the type of the reference (immutable or mutable) does not matter. This means that you can compare an immutable & reference with a mutable one &mut of the same underlying type.

let i = &0;
let m = &mut 1;

i == m; // `false`
m == i; // `false`
m == m; // `true`
i == i; // `true`

The above is equivalent to applying an explicit freeze to each mutable reference where needed

let i = &0;
let m = &mut 1;

i == freeze(m); // `false`
freeze(m) == i; // `false`
m == m; // `true`
i == i; // `true`

But again, the underlying type must be the same type

let i = &0;
let s = &b"";

i == s; // ERROR!
//   ^ expected an argument of type '&u64'

Restrictions

Both == and != consume the value when comparing them. As a result, the type system enforces that the type must have drop. Recall that without the drop ability, ownership must be transferred by the end of the function, and such values can only be explicitly destroyed within their declaring module. If these were used directly with either equality == or non-equality !=, the value would be destroyed which would break drop ability safety guarantees!

address 0x42 {
module example {
    struct Coin has store { value: u64 }
    fun invalid(c1: Coin, c2: Coin) {
        c1 == c2 // ERROR!
//      ^^    ^^ These resources would be destroyed!
    }
}
}

But, a programmer can always borrow the value first instead of directly comparing the value, and reference types have the drop ability. For example

address 0x42 {
module example {
    struct Coin as store { value: u64 }
    fun swap_if_equal(c1: Coin, c2: Coin): (Coin, Coin) {
        let are_equal = &c1 == &c2; // valid
        if (are_equal) (c2, c1) else (c1, c2)
    }
}
}

Avoid Extra Copies

While a programmer can compare any value whose type has drop, a programmer should often compare by reference to avoid expensive copies.

let v1: vector<u8> = function_that_returns_vector();
let v2: vector<u8> = function_that_returns_vector();
assert!(copy v1 == copy v2, 42);
//     ^^^^       ^^^^
use_two_vectors(v1, v2);

let s1: Foo = function_that_returns_large_struct();
let s2: Foo = function_that_returns_large_struct();
assert!(copy s1 == copy s2, 42);
//     ^^^^       ^^^^
use_two_foos(s1, s2);

This code is perfectly acceptable (assuming Foo has drop), just not efficient. The highlighted copies can be removed and replaced with borrows

let v1: vector<u8> = function_that_returns_vector();
let v2: vector<u8> = function_that_returns_vector();
assert!(&v1 == &v2, 42);
//     ^      ^
use_two_vectors(v1, v2);

let s1: Foo = function_that_returns_large_struct();
let s2: Foo = function_that_returns_large_struct();
assert!(&s1 == &s2, 42);
//     ^      ^
use_two_foos(s1, s2);

The efficiency of the == itself remains the same, but the copys are removed and thus the program is more efficient.