On 3/13/12 1:20 PM, Manu wrote:
What value does it add over Kenji's change? Is this because Kenji's change is unable to perform direct to existing variables?
Yes.
My understanding from early in the thread was that Kenji's change hides the returned tuple, and performs a convenient unpack. How can you perform a scatter if the tuple instance is no longer visible?
If I understand you correctly, you just say fun().scatter(v1, v2, v3). Andrei
