Function std::any::type_name

help
1

Hello, is it possible to write an expression in one line to print the type of a variable using the function from the subject?

2

Well, if you really want a one-liner - here it is:

fn main() {
    let var = Some(("s", 1, 1.0));
    let name = {fn tmp<T>(_: T) -> &'static str {std::any::type_name::<T>()}; tmp}(var);
    println!("{}", name);
}

Ugly, but it works - playground.

I'd advise, anyway, not to insist on the one-line solution and to extract the function from the block.

2 Likes
3

I thought it was impossible to write in the form of a simple function call. Thank you, I just started to study Rust, and thought I missed something.

In any case, your code looks very cool. Thanks again!

4

Note this doesn't universally work. It happens to work in your example because the type of var is Copy. The following doesn't compile:

fn main() {
    let var = Some((String::from("s"), 1, 1.0));
    let name = {fn tmp<T>(_: T) -> &'static str {std::any::type_name::<T>()}; tmp}(var);
    println!("{}: {:?}", name, var);
}

(playground)
Because var is moved by the function call.

The following, however, does work in more cases:

fn main() {
    let var = Some((String::from("s"), 1, 1.0));
    let name = {fn tmp<T>(_: &T) -> &'static str {std::any::type_name::<T>()}; tmp}(&var);
    println!("{}: {:?}", name, var);
}

(playground)

2 Likes
5

Thanks, missed the possibility that the function argument and the type parameter might be different.

6

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