Local Variables and Scope
Local variables in Move are lexically (statically) scoped. New variables are introduced with the
keyword let, which will shadow any previous local with the same name. Locals marked as mut are
mutable and can be updated both directly and via a mutable reference.
Declaring Local Variables
let bindings
Move programs use let to bind variable names to values:
let x = 1;
let y = x + x:
let can also be used without binding a value to the local.
let x;
The local can then be assigned a value later.
let x;
if (cond) {
x = 1
} else {
x = 0
}
This can be very helpful when trying to extract a value from a loop when a default value cannot be provided.
let x;
let cond = true;
let i = 0;
loop {
let (res, cond) = foo(i);
if (!cond) {
x = res;
break;
};
i = i + 1;
}
To modify a local variable after it is assigned, or to borrow it mutably &mut, it must be
declared as mut.
let mut x = 0;
if (cond) x = x + 1;
foo(&mut x);
For more details see the section on assignments below.
Variables must be assigned before use
Move's type system prevents a local variable from being used before it has been assigned.
let x;
x + x // ERROR! x is used before being assigned
let x;
if (cond) x = 0;
x + x // ERROR! x does not have a value in all cases
let x;
while (cond) x = 0;
x + x // ERROR! x does not have a value in all cases
Valid variable names
Variable names can contain underscores _, letters a to z, letters A to Z, and digits 0
to 9. Variable names must start with either an underscore _ or a letter a through z. They
cannot start with uppercase letters.
// all valid
let x = e;
let _x = e;
let _A = e;
let x0 = e;
let xA = e;
let foobar_123 = e;
// all invalid
let X = e; // ERROR!
let Foo = e; // ERROR!
Type annotations
The type of a local variable can almost always be inferred by Move's type system. However, Move allows explicit type annotations that can be useful for readability, clarity, or debuggability. The syntax for adding a type annotation is:
let x: T = e; // "Variable x of type T is initialized to expression e"
Some examples of explicit type annotations:
module 0x42::example {
public struct S { f: u64, g: u64 }
fun annotated() {
let u: u8 = 0;
let b: vector<u8> = b"hello";
let a: address = @0x0;
let (x, y): (&u64, &mut u64) = (&0, &mut 1);
let S { f, g: f2 }: S = S { f: 0, g: 1 };
}
}
Note that the type annotations must always be to the right of the pattern:
// ERROR! should be let (x, y): (&u64, &mut u64) = ...
let (x: &u64, y: &mut u64) = (&0, &mut 1);
When annotations are necessary
In some cases, a local type annotation is required if the type system cannot infer the type. This commonly occurs when the type argument for a generic type cannot be inferred. For example:
let _v1 = vector[]; // ERROR!
// ^^^^^^^^ Could not infer this type. Try adding an annotation
let v2: vector<u64> = vector[]; // no error
In a rarer case, the type system might not be able to infer a type for divergent code (where all the
following code is unreachable). Both return and
abort are expressions and can have any type. A
loop has type () if it has a break (or T if has a break e where
e: T), but if there is no break out of the loop, it could have any type. If these types cannot
be inferred, a type annotation is required. For example, this code:
let a: u8 = return ();
let b: bool = abort 0;
let c: signer = loop ();
let x = return (); // ERROR!
// ^ Could not infer this type. Try adding an annotation
let y = abort 0; // ERROR!
// ^ Could not infer this type. Try adding an annotation
let z = loop (); // ERROR!
// ^ Could not infer this type. Try adding an annotation
Adding type annotations to this code will expose other errors about dead code or unused local variables, but the example is still helpful for understanding this problem.
Multiple declarations with tuples
let can introduce more than one local at a time using tuples. The locals declared inside the
parenthesis are initialized to the corresponding values from the tuple.
let () = ();
let (x0, x1) = (0, 1);
let (y0, y1, y2) = (0, 1, 2);
let (z0, z1, z2, z3) = (0, 1, 2, 3);
The type of the expression must match the arity of the tuple pattern exactly.
let (x, y) = (0, 1, 2); // ERROR!
let (x, y, z, q) = (0, 1, 2); // ERROR!
You cannot declare more than one local with the same name in a single let.
let (x, x) = 0; // ERROR!
The mutability of the local variables declared can be mixed.
let (mut x, y) = (0, 1);
x = 1;
Multiple declarations with structs
let can also introduce more than one local at a time when destructuring (or matching against) a
struct. In this form, the let creates a set of local variables that are initialized to the values
of the fields from a struct. The syntax looks like this:
public struct T { f1: u64, f2: u64 }
let T { f1: local1, f2: local2 } = T { f1: 1, f2: 2 };
// local1: u64
// local2: u64
Similarly for positional structs
public struct P(u64, u64)
and
let P(local1, local2) = T { f1: 1, f2: 2 };
// local1: u64
// local2: u64
Here is a more complicated example:
module 0x42::example {
public struct X(u64)
public struct Y { x1: X, x2: X }
fun new_x(): X {
X(1)
}
fun example() {
let Y { x1: X(f), x2 } = Y { x1: new_x(), x2: new_x() };
assert!(f + x2.f == 2, 42);
let Y { x1: X(f1), x2: X(f2) } = Y { x1: new_x(), x2: new_x() };
assert!(f1 + f2 == 2, 42);
}
}
Fields of structs can serve double duty, identifying the field to bind and the name of the variable. This is sometimes referred to as punning.
let Y { x1, x2 } = e;
is equivalent to:
let Y { x1: x1, x2: x2 } = e;
As shown with tuples, you cannot declare more than one local with the same name in a single let.
let Y { x1: x, x2: x } = e; // ERROR!
And as with tuples, the mutability of the local variables declared can be mixed.
let Y { x1: mut x1, x2 } = e;
Furthermore, the mutability of annotation can be applied to the punned fields. Giving the equivalent example
let Y { mut x1, x2 } = e;
Destructuring against references
In the examples above for structs, the bound value in the let was moved, destroying the struct value and binding its fields.
public struct T { f1: u64, f2: u64 }
let T { f1: local1, f2: local2 } = T { f1: 1, f2: 2 };
// local1: u64
// local2: u64
In this scenario the struct value T { f1: 1, f2: 2 } no longer exists after the let.
If you wish instead to not move and destroy the struct value, you can borrow each of its fields. For example:
let t = T { f1: 1, f2: 2 };
let T { f1: local1, f2: local2 } = &t;
// local1: &u64
// local2: &u64
And similarly with mutable references:
let mut t = T { f1: 1, f2: 2 };
let T { f1: local1, f2: local2 } = &mut t;
// local1: &mut u64
// local2: &mut u64
This behavior can also work with nested structs.
module 0x42::example {
public struct X(u64)
public struct Y { x1: X, x2: X }
fun new_x(): X {
X(1)
}
fun example() {
let mut y = Y { x1: new_x(), x2: new_x() };
let Y { x1: X(f), x2 } = &y;
assert!(*f + x2.f == 2, 42);
let Y { x1: X(f1), x2: X(f2) } = &mut y;
*f1 = *f1 + 1;
*f2 = *f2 + 1;
assert!(*f1 + *f2 == 4, 42);
}
}
Ignoring Values
In let bindings, it is often helpful to ignore some values. Local variables that start with _
will be ignored and not introduce a new variable
fun three(): (u64, u64, u64) {
(0, 1, 2)
}
let (x1, _, z1) = three();
let (x2, _y, z2) = three();
assert!(x1 + z1 == x2 + z2, 42);
This can be necessary at times as the compiler will warn on unused local variables
let (x1, y, z1) = three(); // WARNING!
// ^ unused local 'y'
General let grammar
All of the different structures in let can be combined! With that we arrive at this general
grammar for let statements:
let-binding → let pattern-or-list type-annotationopt > initializeropt > pattern-or-list → pattern | ( pattern-list ) > pattern-list → pattern ,opt | pattern , pattern-list > type-annotation → : type initializer → = expression
The general term for the item that introduces the bindings is a pattern. The pattern serves to both destructure data (possibly recursively) and introduce the bindings. The pattern grammar is as follows:
pattern → local-variable | struct-type { field-binding-list } > field-binding-list → field-binding ,opt | field-binding , > field-binding-list > field-binding → field | field : pattern
A few concrete examples with this grammar applied:
let (x, y): (u64, u64) = (0, 1);
// ^ local-variable
// ^ pattern
// ^ local-variable
// ^ pattern
// ^ pattern-list
// ^^^^ pattern-list
// ^^^^^^ pattern-or-list
// ^^^^^^^^^^^^ type-annotation
// ^^^^^^^^ initializer
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ let-binding
let Foo { f, g: x } = Foo { f: 0, g: 1 };
// ^^^ struct-type
// ^ field
// ^ field-binding
// ^ field
// ^ local-variable
// ^ pattern
// ^^^^ field-binding
// ^^^^^^^ field-binding-list
// ^^^^^^^^^^^^^^^ pattern
// ^^^^^^^^^^^^^^^ pattern-or-list
// ^^^^^^^^^^^^^^^^^^^^ initializer
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ let-binding
Mutations
Assignments
After the local is introduced (either by let or as a function parameter), a mut local can be
modified via an assignment:
x = e
Unlike let bindings, assignments are expressions. In some languages, assignments return the value
that was assigned, but in Move, the type of any assignment is always ().
(x = e: ())
Practically, assignments being expressions means that they can be used without adding a new
expression block with braces ({...}).
let x;
if (cond) x = 1 else x = 2;
The assignment uses the similar pattern syntax scheme as let bindings, but with absence of mut:
module 0x42::example {
public struct X { f: u64 }
fun new_x(): X {
X { f: 1 }
}
// Note: this example will complain about unused variables and assignments.
fun example() {
let (mut x, mut y, mut f, mut g) = (0, 0, 0, 0);
(X { f }, X { f: x }) = (new_x(), new_x());
assert!(f + x == 2, 42);
(x, y, f, _, g) = (0, 0, 0, 0, 0);
}
}
Note that a local variable can only have one type, so the type of the local cannot change between assignments.
let mut x;
x = 0;
x = false; // ERROR!
Mutating through a reference
In addition to directly modifying a local with assignment, a mut local can be modified via a
mutable reference &mut.
let mut x = 0;
let r = &mut x;
*r = 1;
assert!(x == 1, 42);
This is particularly useful if either:
(1) You want to modify different variables depending on some condition.
let mut x = 0;
let mut y = 1;
let r = if (cond) &mut x else &mut y;
*r = *r + 1;
(2) You want another function to modify your local value.
let mut x = 0;
modify_ref(&mut x);
This sort of modification is how you modify structs and vectors!
let mut v = vector[];
vector::push_back(&mut v, 100);
assert!(*vector::borrow(&v, 0) == 100, 42);
For more details, see Move references.
Scopes
Any local declared with let is available for any subsequent expression, within that scope.
Scopes are declared with expression blocks, {...}.
Locals cannot be used outside of the declared scope.
let x = 0;
{
let y = 1;
};
x + y // ERROR!
// ^ unbound local 'y'
But, locals from an outer scope can be used in a nested scope.
{
let x = 0;
{
let y = x + 1; // valid
}
}
Locals can be mutated in any scope where they are accessible. That mutation survives with the local, regardless of the scope that performed the mutation.
let mut x = 0;
x = x + 1;
assert!(x == 1, 42);
{
x = x + 1;
assert!(x == 2, 42);
};
assert!(x == 2, 42);
Expression Blocks
An expression block is a series of statements separated by semicolons (;). The resulting value of
an expression block is the value of the last expression in the block.
{ let x = 1; let y = 1; x + y }
In this example, the result of the block is x + y.
A statement can be either a let declaration or an expression. Remember that assignments (x = e)
are expressions of type ().
{ let x; let y = 1; x = 1; x + y }
Function calls are another common expression of type (). Function calls that modify data are
commonly used as statements.
{ let v = vector[]; vector::push_back(&mut v, 1); v }
This is not just limited to () types---any expression can be used as a statement in a sequence!
{
let x = 0;
x + 1; // value is discarded
x + 2; // value is discarded
b"hello"; // value is discarded
}
But! If the expression contains a resource (a value without the drop ability),
you will get an error. This is because Move's type system guarantees that any value that is dropped
has the drop ability. (Ownership must be transferred or the value must be
explicitly destroyed within its declaring module.)
{
let x = 0;
Coin { value: x }; // ERROR!
// ^^^^^^^^^^^^^^^^^ unused value without the `drop` ability
x
}
If a final expression is not present in a block---that is, if there is a trailing semicolon ;,
there is an implicit unit () value. Similarly, if the
expression block is empty, there is an implicit unit () value.
Both are equivalent
{ x = x + 1; 1 / x; }
{ x = x + 1; 1 / x; () }
Similarly both are equivalent
{ }
{ () }
An expression block is itself an expression and can be used anyplace an expression is used. (Note: The body of a function is also an expression block, but the function body cannot be replaced by another expression.)
let my_vector: vector<vector<u8>> = {
let mut v = vector[];
vector::push_back(&mut v, b"hello");
vector::push_back(&mut v, b"goodbye");
v
};
(The type annotation is not needed in this example and only added for clarity.)
Shadowing
If a let introduces a local variable with a name already in scope, that previous variable can no
longer be accessed for the rest of this scope. This is called shadowing.
let x = 0;
assert!(x == 0, 42);
let x = 1; // x is shadowed
assert!(x == 1, 42);
When a local is shadowed, it does not need to retain the same type as before.
let x = 0;
assert!(x == 0, 42);
let x = b"hello"; // x is shadowed
assert!(x == b"hello", 42);
After a local is shadowed, the value stored in the local still exists, but will no longer be
accessible. This is important to keep in mind with values of types without the
drop ability, as ownership of the value must be transferred by the end of the
function.
module 0x42::example {
public struct Coin has store { value: u64 }
fun unused_coin(): Coin {
let x = Coin { value: 0 }; // ERROR!
// ^ This local still contains a value without the `drop` ability
x.value = 1;
let x = Coin { value: 10 };
x
// ^ Invalid return
}
}
When a local is shadowed inside a scope, the shadowing only remains for that scope. The shadowing is gone once that scope ends.
let x = 0;
{
let x = 1;
assert!(x == 1, 42);
};
assert!(x == 0, 42);
Remember, locals can change type when they are shadowed.
let x = 0;
{
let x = b"hello";
assert!(x = b"hello", 42);
};
assert!(x == 0, 42);
Move and Copy
All local variables in Move can be used in two ways, either by move or copy. If one or the other
is not specified, the Move compiler is able to infer whether a copy or a move should be used.
This means that in all of the examples above, a move or a copy would be inserted by the
compiler. A local variable cannot be used without the use of move or copy.
copy will likely feel the most familiar coming from other programming languages, as it creates a
new copy of the value inside of the variable to use in that expression. With copy, the local
variable can be used more than once.
let x = 0;
let y = copy x + 1;
let z = copy x + 2;
Any value with the copy ability can be copied in this way, and will be copied
implicitly unless a move is specified.
move takes the value out of the local variable without copying the data. After a move occurs,
the local variable is unavailable, even if the value's type has the copy
ability.
let x = 1;
let y = move x + 1;
// ------ Local was moved here
let z = move x + 2; // Error!
// ^^^^^^ Invalid usage of local 'x'
y + z
Safety
Move's type system will prevent a value from being used after it is moved. This is the same safety
check described in let declaration that prevents local variables from being used
before it is assigned a value.
Inference
As mentioned above, the Move compiler will infer a copy or move if one is not indicated. The
algorithm for doing so is quite simple:
- Any value with the
copyability is given acopy. - Any reference (both mutable
&mutand immutable&) is given acopy.- Except under special circumstances where it is made a
movefor predictable borrow checker errors. This will happen once the reference is no longer used.
- Except under special circumstances where it is made a
- Any other value is given a
move.
Given the structs
public struct Foo has copy, drop, store { f: u64 }
public struct Coin has store { value: u64 }
we have the following example
let s = b"hello";
let foo = Foo { f: 0 };
let coin = Coin { value: 0 };
let coins = vector[Coin { value: 0 }, Coin { value: 0 }];
let s2 = s; // copy
let foo2 = foo; // copy
let coin2 = coin; // move
let coins2 = coin; // move
let x = 0;
let b = false;
let addr = @0x42;
let x_ref = &x;
let coin_ref = &mut coin2;
let x2 = x; // copy
let b2 = b; // copy
let addr2 = @0x42; // copy
let x_ref2 = x_ref; // copy
let coin_ref2 = coin_ref; // copy