Generics
Generics can be used to define functions and structs over different input data types. This language feature is sometimes referred to as parametric polymorphism. In Move, we will often use the term generics interchangeably with type parameters and type arguments.
Generics are commonly used in library code, such as in vector, to declare code that works over any possible type (that satisfies the specified constraints). This sort of parameterization allows you to reuse the same implementation across multiple types and situations.
Declaring Type Parameters
Both functions and structs can take a list of type parameters in their signatures, enclosed by a
pair of angle brackets <...>
.
Generic Functions
Type parameters for functions are placed after the function name and before the (value) parameter list. The following code defines a generic identity function that takes a value of any type and returns that value unchanged.
fun id<T>(x: T): T {
// this type annotation is unnecessary but valid
(x: T)
}
Once defined, the type parameter T
can be used in parameter types, return types, and inside the
function body.
Generic Structs
Type parameters for structs are placed after the struct name, and can be used to name the types of the fields.
public struct Foo<T> has copy, drop { x: T }
public struct Bar<T1, T2> has copy, drop {
x: T1,
y: vector<T2>,
}
Note that type parameters do not have to be used
Type Arguments
Calling Generic Functions
When calling a generic function, one can specify the type arguments for the function's type parameters in a list enclosed by a pair of angle brackets.
fun foo() {
let x = id<bool>(true);
}
If you do not specify the type arguments, Move's type inference will supply them for you.
Using Generic Structs
Similarly, one can attach a list of type arguments for the struct's type parameters when constructing or destructing values of generic types.
fun foo() {
// type arguments on construction
let foo = Foo<bool> { x: true };
let bar = Bar<u64, u8> { x: 0, y: vector<u8>[] };
// type arguments on destruction
let Foo<bool> { x } = foo;
let Bar<u64, u8> { x, y } = bar;
}
In any case if you do not specify the type arguments, Move's type inference will supply them for you.
Type Argument Mismatch
If you specify the type arguments and they conflict with the actual values supplied, an error will be given:
fun foo() {
let x = id<u64>(true); // ERROR! true is not a u64
}
and similarly:
fun foo() {
let foo = Foo<bool> { x: 0 }; // ERROR! 0 is not a bool
let Foo<address> { x } = foo; // ERROR! bool is incompatible with address
}
Type Inference
In most cases, the Move compiler will be able to infer the type arguments so you don't have to write them down explicitly. Here's what the examples above would look like if we omit the type arguments:
fun foo() {
let x = id(true);
// ^ <bool> is inferred
let foo = Foo { x: true };
// ^ <bool> is inferred
let Foo { x } = foo;
// ^ <bool> is inferred
}
Note: when the compiler is unable to infer the types, you'll need annotate them manually. A common scenario is to call a function with type parameters appearing only at return positions.
module a::m {
fun foo() {
let v = vector[]; // ERROR!
// ^ The compiler cannot figure out the element type, since it is never used
let v = vector<u64>[];
// ^~~~~ Must annotate manually in this case.
}
}
Note that these cases are a bit contrived since the vector[]
is never used, ad as such, Move's
type inference cannot infer the type.
However, the compiler will be able to infer the type if that value is used later in that function:
module a::m {
fun foo() {
let v = vector[];
// ^ <u64> is inferred
vector::push_back(&mut v, 42);
// ^ <u64> is inferred
}
}
_
Type
In some cases, you might want to explicitly annotate some of the type arguments, but let the
compiler infer the others. The _
type serves as such a placeholder for the compiler to infer the
type.
let bar = Bar<u64, _> { x: 0, y: vector[b"hello"] };
// ^ vector<u8> is inferred
The placeholder _
may only appear in expressions and macro function definitions, not signatures.
This means you cannot use _
as part of the definition of a function parameter, function return
type, constant definition type, and datatype field.
Integers
In Move, the integer types u8
, u16
, u32
, u64
, u128
, and u256
are all distinct types.
However, each one of these types can be created with the same numerical value syntax. In other
words, if a type suffix is not provided, the compiler will infer the integer type based on the usage
of the value.
let x8: u8 = 0;
let x16: u16 = 0;
let x32: u32 = 0;
let x64: u64 = 0;
let x128: u128 = 0;
let x256: u256 = 0;
If the value is not used in a context that requires a specific integer type, u64
is taken as a
default.
let x = 0;
// ^ u64 is used by default
If the value however is too large for the inferred type, an error will be given
let i: u8 = 256; // ERROR!
// ^^^ too large for u8
let x = 340282366920938463463374607431768211454;
// ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ too large for u64
In cases where the number is too large, you might need to annotate it explicitly
let x = 340282366920938463463374607431768211454u128;
// ^^^^ valid!
Unused Type Parameters
For a struct definition, an unused type parameter is one that does not appear in any field defined in the struct, but is checked statically at compile time. Move allows unused type parameters so the following struct definition is valid:
public struct Foo<T> {
foo: u64
}
This can be convenient when modeling certain concepts. Here is an example:
module a::m {
// Currency Specifiers
public struct A {}
public struct B {}
// A generic coin type that can be instantiated using a currency
// specifier type.
// e.g. Coin<A>, Coin<B> etc.
public struct Coin<Currency> has store {
value: u64
}
// Write code generically about all currencies
public fun mint_generic<Currency>(value: u64): Coin<Currency> {
Coin { value }
}
// Write code concretely about one currency
public fun mint_a(value: u64): Coin<A> {
mint_generic(value)
}
public fun mint_b(value: u64): Coin<B> {
mint_generic(value)
}
}
In this example, Coin<Currency>
is generic on the Currency
type parameter, which specifies the
currency of the coin and allows code to be written either generically on any currency or concretely
on a specific currency. This generality applies even when the Currency
type parameter does not
appear in any of the fields defined in Coin
.
Phantom Type Parameters
In the example above, although struct Coin
asks for the store
ability, neither Coin<A>
nor
Coin<B>
will have the store
ability. This is because of the rules for
Conditional Abilities and Generic Types
and the fact that A
and B
don't have the store
ability, despite the fact that they are not
even used in the body of struct Coin
. This might cause some unpleasant consequences. For example,
we are unable to put Coin<A>
into a wallet in storage.
One possible solution would be to add spurious ability annotations to A
and B
(i.e.,
public struct Currency1 has store {}
). But, this might lead to bugs or security vulnerabilities
because it weakens the types with unnecessary ability declarations. For example, we would never
expect a value in the storage to have a field in type A
, but this would be possible with the
spurious store
ability. Moreover, the spurious annotations would be infectious, requiring many
functions generic on the unused type parameter to also include the necessary constraints.
Phantom type parameters solve this problem. Unused type parameters can be marked as phantom type
parameters, which do not participate in the ability derivation for structs. In this way, arguments
to phantom type parameters are not considered when deriving the abilities for generic types, thus
avoiding the need for spurious ability annotations. For this relaxed rule to be sound, Move's type
system guarantees that a parameter declared as phantom
is either not used at all in the struct
definition, or it is only used as an argument to type parameters also declared as phantom
.
Declaration
In a struct definition a type parameter can be declared as phantom by adding the phantom
keyword
before its declaration.
public struct Coin<phantom Currency> has store {
value: u64
}
If a type parameter is declared as phantom we say it is a phantom type parameter. When defining a struct, Move's type checker ensures that every phantom type parameter is either not used inside the struct definition or it is only used as an argument to a phantom type parameter.
public struct S1<phantom T1, T2> { f: u64 }
// ^^^^^^^ valid, T1 does not appear inside the struct definition
public struct S2<phantom T1, T2> { f: S1<T1, T2> }
// ^^^^^^^ valid, T1 appears in phantom position
The following code shows examples of violations of the rule:
public struct S1<phantom T> { f: T }
// ^^^^^^^ ERROR! ^ Not a phantom position
public struct S2<T> { f: T }
public struct S3<phantom T> { f: S2<T> }
// ^^^^^^^ ERROR! ^ Not a phantom position
More formally, if a type is used as an argument to a phantom type parameter we say the type appears in phantom position. With this definition in place, the rule for the correct use of phantom parameters can be specified as follows: A phantom type parameter can only appear in phantom position.
Note that specifying phantom
is not required, but the compiler will warn if a type parameter could
be phantom
but was not marked as such.
Instantiation
When instantiating a struct, the arguments to phantom parameters are excluded when deriving the struct abilities. For example, consider the following code:
public struct S<T1, phantom T2> has copy { f: T1 }
public struct NoCopy {}
public struct HasCopy has copy {}
Consider now the type S<HasCopy, NoCopy>
. Since S
is defined with copy
and all non-phantom
arguments have copy
then S<HasCopy, NoCopy>
also has copy
.
Phantom Type Parameters with Ability Constraints
Ability constraints and phantom type parameters are orthogonal features in the sense that phantom parameters can be declared with ability constraints.
public struct S<phantom T: copy> {}
When instantiating a phantom type parameter with an ability constraint, the type argument has to
satisfy that constraint, even though the parameter is phantom. The usual restrictions apply and T
can only be instantiated with arguments having copy
.
Constraints
In the examples above, we have demonstrated how one can use type parameters to define "unknown" types that can be plugged in by callers at a later time. This however means the type system has little information about the type and has to perform checks in a very conservative way. In some sense, the type system must assume the worst case scenario for an unconstrained generic--a type with no abilities.
Constraints offer a way to specify what properties these unknown types have so the type system can allow operations that would otherwise be unsafe.
Declaring Constraints
Constraints can be imposed on type parameters using the following syntax.
// T is the name of the type parameter
T: <ability> (+ <ability>)*
The <ability>
can be any of the four abilities, and a type parameter can be
constrained with multiple abilities at once. So all of the following would be valid type parameter
declarations:
T: copy
T: copy + drop
T: copy + drop + store + key
Verifying Constraints
Constraints are checked at instantiation sites
public struct Foo<T: copy> { x: T }
public struct Bar { x: Foo<u8> }
// ^^ valid, u8 has `copy`
public struct Baz<T> { x: Foo<T> }
// ^ ERROR! T does not have 'copy'
And similarly for functions
fun unsafe_consume<T>(x: T) {
// ERROR! x does not have 'drop'
}
fun consume<T: drop>(x: T) {
// valid, x will be dropped automatically
}
public struct NoAbilities {}
fun foo() {
let r = NoAbilities {};
consume<NoAbilities>(NoAbilities);
// ^^^^^^^^^^^ ERROR! NoAbilities does not have 'drop'
}
And some similar examples, but with copy
fun unsafe_double<T>(x: T) {
(copy x, x)
// ERROR! T does not have 'copy'
}
fun double<T: copy>(x: T) {
(copy x, x) // valid, T has 'copy'
}
public struct NoAbilities {}
fun foo(): (NoAbilities, NoAbilities) {
let r = NoAbilities {};
double<NoAbilities>(r)
// ^ ERROR! NoAbilities does not have 'copy'
}
For more information, see the abilities section on conditional abilities and generic types.
Limitations on Recursions
Recursive Structs
Generic structs can not contain fields of the same type, either directly or indirectly, even with different type arguments. All of the following struct definitions are invalid:
public struct Foo<T> {
x: Foo<u64> // ERROR! 'Foo' containing 'Foo'
}
public struct Bar<T> {
x: Bar<T> // ERROR! 'Bar' containing 'Bar'
}
// ERROR! 'A' and 'B' forming a cycle, which is not allowed either.
public struct A<T> {
x: B<T, u64>
}
public struct B<T1, T2> {
x: A<T1>
y: A<T2>
}
Advanced Topic: Type-level Recursions
Move allows generic functions to be called recursively. However, when used in combination with generic structs, this could create an infinite number of types in certain cases, and allowing this means adding unnecessary complexity to the compiler, vm and other language components. Therefore, such recursions are forbidden.
This restriction might be relaxed in the future, but for now, the following examples should give you an idea of what is allowed and what is not.
module a::m {
public struct A<T> {}
// Finitely many types -- allowed.
// foo<T> -> foo<T> -> foo<T> -> ... is valid
fun foo<T>() {
foo<T>();
}
// Finitely many types -- allowed.
// foo<T> -> foo<A<u64>> -> foo<A<u64>> -> ... is valid
fun foo<T>() {
foo<A<u64>>();
}
}
Not allowed:
module a::m {
public struct A<T> {}
// Infinitely many types -- NOT allowed.
// error!
// foo<T> -> foo<A<T>> -> foo<A<A<T>>> -> ...
fun foo<T>() {
foo<Foo<T>>();
}
}
And similarly, not allowed:
module a::n {
public struct A<T> {}
// Infinitely many types -- NOT allowed.
// error!
// foo<T1, T2> -> bar<T2, T1> -> foo<T2, A<T1>>
// -> bar<A<T1>, T2> -> foo<A<T1>, A<T2>>
// -> bar<A<T2>, A<T1>> -> foo<A<T2>, A<A<T1>>>
// -> ...
fun foo<T1, T2>() {
bar<T2, T1>();
}
fun bar<T1, T2> {
foo<T1, A<T2>>();
}
}
Note, the check for type level recursions is based on a conservative analysis on the call sites and does NOT take control flow or runtime values into account.
module a::m {
public struct A<T> {}
// Infinitely many types -- NOT allowed.
// error!
fun foo<T>(n: u64) {
if (n > 0) foo<A<T>>(n - 1);
}
}
The function in the example above will technically terminate for any given input and therefore only creating finitely many types, but it is still considered invalid by Move's type system.