I want to initialize an array whose length is equal to another array's length:
fn foo(array: &[i32]) {
let mut sum = [0; array.len()];
}
It will make an error:
error[E0080]: constant evaluation error
--> test.rs:22:18
|
22 | let mut sum = [0; array.len()];
| ^^^^^^^^^^^ unsupported constant expr
I think I must use this len() argument... How can I solve this?
To answer the question asked:
How to use another array's length to initialize an array in Rust?
As of Rust 1.51, you can write generic code like:
fn foo<const N: usize>(array: [u8; N]) -> [u8; N] {
unimplemented!()
}
See Is it possible to control the size of an array using the type parameter of a generic? for further details.
The solution for previous versions of Rust involves a combination of traits and macros. This has the distinct disadvantage of being limited to a set of types. This is discussed further in questions like Why does println! work only for arrays with a length less than 33?
Showcasing the shorter, more idiomatic version of the Vec-based code:
fn foo(slice: &[i32]) {
let mut sum = vec![0; slice.len()];
}
You are actually taking the length of a slice, not an array. Array lengths must be known at compile time. array.len(), being the length of a slice, is potentially derived from runtime input, and as such cannot be used as an array length. You can create a vector instead. For example:
use std::iter::repeat;
fn foo(slice: &[i32]){
let mut sum: Vec<i32> = repeat(0).take(slice.len()).collect();
// ......
}
vec![0; slice.len()].