Equality

Move supports two equality operations == and !=

Operations

Syntax	Operation	Description
==	equal	Returns true if the two operands have the same value, false otherwise
!=	not equal	Returns true if the two operands have different values, false otherwise

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 { 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 does not matter. This means that you can compare an immutable & reference with a mutable one &mut of the same 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 similar to non-reference types, 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 be copyable, that is that it is not a resource value. Recall that resources cannot be copied, ownership must be transferred by the end of the function, and they can only be explicitly destroyed within their declaring module. If resources were used directly with either equality == or non-equality !=, the value would be destroyed which would break resource safety!

address 0x42 {

module Example {

resource struct Coin { 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--to make it a copyable type--instead of directly comparing the resource. For example

address 0x42 {

module Example {

resource struct Coin { 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 copyable value, 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, 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.

tags: basics Reviewed by Legal