How to set a Rust array length dynamically?

Is it possible to create array with dynamic length?

No. By definition, arrays have a length defined at compile time. A variable (because it can vary) is not known at compile time. The compiler would not know how much space to allocate on the stack to provide storage for the array.

You will need to use a Vec:

let arr = vec![0; length];

See also:

• Is it possible to control the size of an array using the type parameter of a generic?

You can create your own HeapArray. It's not that complicated if you read alloc's docs:

use std::alloc::{alloc, dealloc, Layout};pub struct HeapArray<T> {    ptr: *mut T,    len: usize,}impl<T> HeapArray<T> {    pub fn new(len: usize) -> Self {        let ptr = unsafe {            let layout = Layout::from_size_align_unchecked(len, std::mem::size_of::<T>());            alloc(layout) as *mut T        };        Self { ptr, len }    }    pub fn get(&self, idx: usize) -> Option<&T> {        if idx < self.len {            unsafe { Some(&*(self.ptr.add(idx))) }        } else {            None        }    }    pub fn get_mut(&self, idx: usize) -> Option<&mut T> {        if idx < self.len {            unsafe { Some(&mut *(self.ptr.add(idx))) }        } else {            None        }    }    pub fn len(&self) -> usize {        self.len    }}impl<T> Drop for HeapArray<T> {    fn drop(&mut self) {        unsafe {            dealloc(                self.ptr as *mut u8,                Layout::from_size_align_unchecked(self.len, std::mem::size_of::<T>()),            )        };    }}impl<T> std::ops::Index<usize> for HeapArray<T> {    type Output = T;    fn index(&self, index: usize) -> &Self::Output {        self.get(index).unwrap()    }}impl<T> std::ops::IndexMut<usize> for HeapArray<T> {    fn index_mut(&mut self, index: usize) -> &mut Self::Output {        self.get_mut(index).unwrap()    }}

You may also add methods like as_slice, get_unchecked, etc.

This should be possible after variable length arrays (VLA) are implemented.