I saw this logic and wonder why the result is one rather than zero:
P ← 1 1 0
Q ← 0 0 1
∨/ P ∧← Q
1
On the other hand the following results in zero:
P ← 1 1 0
Q ← 0 0 1
∨/ P ← P ∧ Q
0
I'm using Dyalog APL 16.
The result of any assignment is the value to the right of the assignment arrow, so P ∧← Q returns Q while P ← P ∧ Q returns P ∧ Q which is also the new value assigned to P, per the documentation:
Assignment (Modified)
{R}←Xf←Y(…)
Ris the “pass-through” value, that is, the value ofY. If the result of the derived function is not assigned or used, there is no explicit result.
To get the performance benefit of in-place modification, while using the new value inline, you can write ∨/ P ⊣ P ∧← Q