First, clarify the code snippet and the exact dispatch mechanism (e.g., DispatchQueue.main.async). Then, reason step-by-step about the execution order: the async blocks are enqueued and executed later on the main queue, so the print statements inside them occur after the current synchronous code. Finally, explain that the second closure prints '123' because it captures a reference to the same mutable String variable, and by the time it runs, the variable has been mutated to '123'.
Pro tip: Mention that this behavior is due to reference capture of variables in closures, not value semantics of String itself. Also, note that if the variable were a let constant or if the closure captured a copy (e.g., via capture list), the output would differ.
Recognize that DispatchQueue.main.async enqueues the closure to run later on the main queue, after the current synchronous code completes.
Since both async blocks are enqueued, they will execute in FIFO order after the current run loop iteration. Any synchronous print statements outside the blocks will execute first.
The closures capture the variable (not its value) by reference. So both closures see the same mutable storage. The second closure sees the final mutated value.
Even though String is a value type, the variable itself is captured by reference. When the second closure runs, the variable has been changed to '123', so it prints '123' instead of '13'.
Clarify that value type semantics apply to the String instance, but the closure captures the variable (a reference to the storage), not a copy of the String. This is why mutation is visible.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.