Copy without losing the source on failure

copy(value) returns Result[T, AllocError]. It observes the source, so your original value remains usable whether copying succeeds or fails. Use ? to keep the success path short:

def extended(source: &list[i64]) -> Result[list[i64], AllocError]:
    mut result = copy(source)?
    result.append(30)?
    Ok(result)

def main() -> Result[(), AllocError | IoError]:
    original = [10, 20]?
    print(extended(&original)?)?
    print(original)?
    Ok(())

[10, 20, 30]
[10, 20]

This also works for nested collections, ordinary classes, and enum payloads. If copying a later field or element fails, the partial copy is cleaned up and the original remains unchanged. Scalars and immutable values, including strings, need no allocation to copy; their existing immutable storage can be shared.

Generators and classes with custom cleanup cannot be copied, including when nested in another value. The operation covers duplication itself. In copy([1, 2]?)?, the first ? handles construction and the second handles copying.