Generics

Generic functions

Throughout this tutorial, I've been defining functions like map and for_rev to take vectors of integers. It is 2011, and we no longer expect to be defining such functions again and again for every type they apply to. Thus, Rust allows functions and datatypes to have type parameters.

fn for_rev<T>(v: [T], act: block(T)) {
    let i = std::vec::len(v);
    while i > 0u {
        i -= 1u;
        act(v[i]);
    }
}

fn map<T, U>(f: block(T) -> U, v: [T]) -> [U] {
    let acc = [];
    for elt in v { acc += [f(elt)]; }
    ret acc;
}

When defined in this way, these functions can be applied to any type of vector, as long as the type of the block's argument and the type of the vector's content agree with each other.

Inside a parameterized (generic) function, the names of the type parameters (capitalized by convention) stand for opaque types. You can't look inside them, but you can pass them around.

Generic datatypes

Generic type and tag declarations follow the same pattern:

type circular_buf<T> = {start: uint,
                        end: uint,
                        buf: [mutable T]};

tag option<T> { some(T); none; }

You can then declare a function to take a circular_buf<u8> or return an option<str>, or even an option<T> if the function itself is generic.

The option type given above exists in the standard library as std::option::t, and is the way Rust programs express the thing that in C would be a nullable pointer. The nice part is that you have to explicitly unpack an option type, so accidental null pointer dereferences become impossible.

Type-inference and generics

Rust's type inferrer works very well with generics, but there are programs that just can't be typed.

let n = std::option::none;
# n = std::option::some(1);

If you never do anything else with n, the compiler will not be able to assign a type to it. (The same goes for [], in fact.) If you really want to have such a statement, you'll have to write it like this:

let n2: std::option::t<int> = std::option::none;
// or
let n = std::option::none::<int>;

Note that, in a value expression, < already has a meaning as a comparison operator, so you'll have to write ::<T> to explicitly give a type to a name that denotes a generic value. Fortunately, this is rarely necessary.

Polymorphic built-ins

There are two built-in operations that, perhaps surprisingly, act on values of any type. It was already mentioned earlier that log can take any type of value and output it as a string.

More interesting is that Rust also defines an ordering for values of all datatypes, and allows you to meaningfully apply comparison operators (<, >, <=, >=, ==, !=) to them. For structural types, the comparison happens left to right, so "abc" < "bac" (but note that "bac" < "ác", because the ordering acts on UTF-8 sequences without any sophistication).

Kinds

Perhaps surprisingly, the 'copy' (duplicate) operation is not defined for all Rust types. Resource types (types with destructors) can not be copied, and neither can any type whose copying would require copying a resource (such as records or unique boxes containing a resource).

This complicates handling of generic functions. If you have a type parameter T, can you copy values of that type? In Rust, you can't, unless you explicitly declare that type parameter to have copyable 'kind'. A kind is a type of type.

// This does not compile
fn head_bad<T>(v: [T]) -> T { v[0] }
// This does
fn head<T:copy>(v: [T]) -> T { v[0] }

When instantiating a generic function, you can only instantiate it with types that fit its kinds. So you could not apply head to a resource type.

Rust has three kinds: 'noncopyable', 'copyable', and 'sendable'. By default, type parameters are considered to be noncopyable. You can annotate them with the copy keyword to declare them copyable, and with the send keyword to make them sendable.

Sendable types are a subset of copyable types. They are types that do not contain shared (reference counted) types, which are thus uniquely owned by the function that owns them, and can be sent over channels to other tasks. Most of the generic functions in the std::comm module take sendable types.

Generic functions and argument-passing

If you try this program:

# fn map(f: block(int) -> int, v: [int]) {}
fn plus1(x: int) -> int { x + 1 }
map(plus1, [1, 2, 3]);

You will get an error message about argument passing styles disagreeing. The reason is that generic types are always passed by pointer, so map expects a function that takes its argument by pointer. The plus1 you defined, however, uses the default, efficient way to pass integers, which is by value. To get around this issue, you have to explicitly mark the arguments to a function that you want to pass to a generic higher-order function as being passed by pointer:

# fn map<T, U>(f: block(T) -> U, v: [T]) {}
fn plus1(&&x: int) -> int { x + 1 }
map(plus1, [1, 2, 3]);

NOTE: This is inconvenient, and we are hoping to get rid of this restriction in the future.

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