Update a set by transferring another set

Sets also support update(source) -> Result[(), AllocError]. Members already present are kept, and missing members transfer from the source:

def main() -> Result[(), AllocError | IoError]:
    mut names = {"Ada", "Bea"}?
    more = {"Bea", "Cam"}?
    names.update(more)?
    print(len(names))?
    print("Ada" in names and "Bea" in names and "Cam" in names)?
    Ok(())

3
True

The source must be a set of the same type and is consumed on both outcomes. Use names.update(copy(source)?) to preserve it. Failed reservation leaves the destination unchanged and cleans up the consumed source. Duplicate-only and empty sources need no new runtime storage; existing capacity can cover new members too. Set iteration order remains unspecified.