How do I pass one type to a generic defined with two types in Rust?

Question

Supposing I have a function like this:

fn my<T: fmt::Debug, U>(x: T) {
    println!("{:?}", mem::size_of::<U>());
    println!("{:?}", x);
}

It depends on two types T and U. I have to specify the type U when I call the function my. On the other hand, the type T can be determined from passed x, so, it's logical, I can miss it in a call.

When I try:

my::<u8>(10);

I get the error:

error[E0107]: wrong number of type arguments: expected 2, found 1

What is the correct way to say to Rust compiler to get the type T from passed x and to pass U only in <...>?

mcarton · Accepted Answer · 2020-05-02 10:48:11Z

4

In Rust, _ in a type position is used to ask the compiler to infer something:

let foo: Vec<_> = (1..10).map(|i| i.to_string()).collect();
//           ^ collect in a vector whose type should be obvious

This also works in your case:

my::<_, u8>(10);
//   ^ infer `T`

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answered May 2, 2020 at 10:48

mcarton

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