rust - 从具有特征的泛型函数返回类型

Question

I'm trying to understand how generic functions works in Rust. Here's my code:

fn evaluate<T: std::ops::Mul>(portfolio: T, quote: T) -> T {
    portfolio * quote
}

And I getting the error:

error[E0308]: mismatched types
  --> src/main.rs:10:5
   |
9  | fn evaluate<T: std::ops::Mul>(portfolio: T, quote: T) -> T {
   |                                                          - expected `T` because of return type
10 |     portfolio * quote
   |     ^^^^^^^^^^^^^^^^^ expected type parameter `T`, found associated type
   |
   = note: expected type parameter `T`
             found associated type `<T as std::ops::Mul>::Output`
   = note: you might be missing a type parameter or trait bound

What i'm doing wrong?

Answer 1

Rust does not assume that multiplication of to values of type T always results in a new value of type T . Instead, the Mul trait has an associated type Output that specifies the return type of the multiplication. The error message mentions that type as "found associated type <T as std::ops::Mul>::Output ". So the easiest solution is to change the return type of your function to the assoicated type:

fn evaluate<T: std::ops::Mul>(portfolio: T, quote: T) -> <T as std::ops::Mul>::Output {
    portfolio * quote
}

A common example where Output does not match the input type is multiplying two references – the output will usually be an owned type, not a reference again.