CODEWITHSUNDEEP
rust
Gunty
Gunty

Reputation: 2479

Rust how to make function accept list of size n?

fn function_name(the_list: ?) {
    // do something
    println!("{}",the_list);
}

fn main() {
    let data = [5,20,18,30,9,7,12,40]; // where data size is n
    let result = function_name(&data);
}

I have a very simple function here which I want to pass a list of size n into. I have looked through the docs and only found solutions for lists whose size is known which is not what I want.

How do I approach this problem?

Upvotes: 0

Views: 445

Answers (2)

Aivean
Aivean

Reputation: 10882

To get compile time information about the size of passed array you can use const generics, like this:

fn function_name<T:Debug, const N: usize>(the_list: &[T; N]) {
    // do something
    println!("{:?}", the_list);
}

fn main() {
    let data = [5,20,18,30,9,7,12,40]; // where data size is n
    let result = function_name(&data);
}

If you want to restrict this size to particular value, you can do this:

fn function_name<T:Debug>(the_list: &[T; 8]) {

P.S., [i32] is called primitive array.

Upvotes: 4

Gunty
Gunty

Reputation: 2479

fn function_name(the_list: &[i32]) {
    // do something
    println!("{:?}",the_list);
}

fn main() {
    let data = [5,20,18,30,9,7,12,40]; // where data size is n
    let result = function_name(&data);
}

//output: [5, 20, 18, 30, 9, 7, 12, 40]

The solution is to have the function accept a general [i32] list type instead. It is not necessary to write the length of the list's input if the input is of size n.

Upvotes: 3

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