struct Foo<T>(T);
And you create Foo(42i32) and Foo(0.0f64), the compiler will create the equivalent to struct Fooi32(i32);
struct Foof64(f64);
In other languages like Java, generics are implemented the way that Rust does "trait objects" (&dyn Trait).Rust is not the only language that does this, to be clear.
If you're interested in a quick intro on the compiler side of this, you can read https://rustc-dev-guide.rust-lang.org/backend/monomorph.html
package main
import (
"fmt"
)
type wrapper[T any] struct {
Value T
}
func (w wrapper[T]) String() string {
return fmt.Sprintf("{%v}", w.Value)
}
func stringWrapped[T any](n int, v T) string {
if n == 0 {
return fmt.Sprintf("%v", v)
}
return stringWrapped(n-1, wrapper[T]{Value: v})
}
func main() {
n := 0
fmt.Scanf("%d", &n)
result := stringWrapped(n, "test")
fmt.Println(result)
}
Go refuses to compile because it can't possibly generate all instances of wrapper[T] that this program may use: wrapper[string], wrapper[wrapper[string]], wrapper[wrapper[wrapper[string]]], etc.[1]: https://play.rust-lang.org/?version=stable&mode=debug&editio...
[2]: https://play.rust-lang.org/?version=stable&mode=debug&editio...
^ This blows the stack because it keeps calling itself with no break condition, but shows how the type system accepted the code.
In C++ you can monomorphize as long as you can somehow prove the recursion terminates at compile time (for example by threading a static recursion counter).
And now the thing is: with transparent signature ascriptions, functors are monomorphised in SML, instead of everything being hidden behind signatures (as is in the case of Rust with traits when you use dyn), which has semantic consequences. E.g. a struct returned by a functor may contain a type. You can't perform proper type-checking without monomorphising, because you don't know what the exact type is. E.g. in the following program, the final line couldn't be type-checked without monomorphisation:
signature ITERABLE = sig
type ElemT
type SrcT
val new_iter: SrcT -> unit -> ElemT option
end
signature LIST_ELEM_TYPE = sig
type T
end
functor ListIterFun (ListElemType: LIST_ELEM_TYPE): ITERABLE = struct
type ElemT = ListElemType.T
type SrcT = ElemT list
fun new_iter l = let val lr = ref l
in
fn () => case !lr of
nil => NONE
| (x::xs) => (lr := xs; SOME x)
end
end
structure IntElemType: LIST_ELEM_TYPE = struct
type T = int
end
structure IntListIter = ListIterFun(IntElemType)
val next = IntListIter.new_iter [1, 2, 3, 4, 5]
If I change the signature ascription on ListIterFun to an opaque ascription (:> ITERABLE), the final line won't type-check, because it's not obvious from the signature, that ElemT is int. So transparent signature ascriptions require monomorphisation (Rust traits without dyn), and opaque signature ascriptions free the compiler from having to do monomorphisation (Rust traits with dyn*).There was a lot of discussion of this issue when Go was settling on a design for its generics, under the phrase "reified generics".
It's probably exactly how templates work, except the details are invisible to users.
https://hacks.mozilla.org/2020/11/warp-improved-js-performan...
Given
fn foo<T>(a: T, b: T) -> T { a + b }
The compiler will complain that you should have been explicit on how T is going to be used: error[E0369]: cannot add `T` to `T`
--> src/lib.rs:1:32
|
1 | fn foo<T>(a: T, b: T) -> T { a + b }
| - ^ - T
| |
| T
|
help: consider restricting type parameter `T`
|
1 | fn foo<T: Add<Output = T>>(a: T, b: T) -> T { a + b }
| +++++++++++++++++
whereas in C++ this would have been accepted until you called foo with two things that couldn't be added together, like a Rust macro[1].[1]: https://play.rust-lang.org/?version=nightly&mode=debug&editi...