Can not infer type of generic function in recursive enums in Rust

Question

I'm working with recursive enums in Rust. I have the following code:

enum Rdd<T, G, H> where G: (Fn(&T) -> H), { Data(Vec<T>), Map(G, Box<Rdd<T, G, H>>) } fn main() { let mut v1: Vec<u32> = vec![1, 2, 3, 4, 5, 6]; let rdd_1 = Rdd::Data(v1); // It does not work }

When I try to compile It throws:

let rdd_1 = Rdd::Data(v1); // It does not work ^^^^^^^^^ cannot infer type for `G` consider giving `rdd_1` the explicit type `Rdd<u32, G, H>`, where the type parameter `G` is specified

Why should I provide a type for G parameter as It's not needed for the Rdd::Data enum? How could I solve this problem?

Thanks in advance

Boiethios · Accepted Answer · 2019-12-02 16:45:13Z

The compiler needs to know all the generic parameters since you could update the value to Rdd::Map, and thus it wants to know its size.

In this situation, I would create my own constructor with dummy default generic parameters:

enum Rdd<T, G = fn(&T) -> (), H = ()> where G: (Fn(&T) -> H), { Data(Vec<T>), Map(G, Box<Rdd<T, G, H>>) } impl<T> Rdd<T, fn(&T), ()> { // for example fn new_data(data: Vec<T>) -> Self { Rdd::Data(data) } } fn main() { let mut v1: Vec<u32> = vec![1, 2, 3, 4, 5, 6]; let rdd_1 = Rdd::new_data(v1); }

Note that you cannot update your rdd_1 with any Map you want.

Thank you soy much! It works. Unfortunatelly as you say, I cannot use any map function, I'll read about GADT in Rust. I don't understand why G is equal to () and then you use the where, but I will finish the book and I hope the explanation is there. Thank you again
new_data creates an Rdd with dummy parameters G and H: that's why I wrote that this solution is useful only when you won't user the Map variant after that.

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