如何重新导出具有具体类型的泛型函数？

Question

I have a generic function foo<T>() . However, I want to specialize the function for the type usize . I could declare another function and call the generic function. However, I am wondering if there is any syntax sugar I could use, such as (pseudo code) pub use foo::<usize> as foo_usize .

use std::fmt::Debug;

fn foo<T: Debug>(a: &T) {
    println!("{:?}", a)
}

// I do not want generics, as I need to export this function as extern with #[no_mangle]
#[no_mangle]
pub extern "C" fn foo_usize(a: &usize) {
    foo::<usize>(a)
}

Answer 1

You will need to perform the monomorphization yourself by explicitly listing out the function and performing the call:

use std::fmt::Debug;

fn foo<T: Debug>(a: &T) {
    println!("{:?}", a)
}

#[no_mangle]
pub extern "C" fn foo_usize(a: &usize) {
    foo::<usize>(a)
}