This follows the template I gave in a previous message.
The first evaluation of u produces a result that is different in
type/shape from the original y. The computation restarts from scratch,
executing u twice more.
In your other examples the result of u was concatenated only with other
results of u, and there was no change in type/shape.
Henry Rich
On 3/5/2020 8:15 PM, bill lam wrote:
How to explain the output of last one which contains 0 in vec ?
empty@:(echo bind 99)^:2 3 1] 9
99
99
99
empty@:(echo bind 99)^:4 3 1] 9
99
99
99
99
empty@:(echo bind 99)^:2 0 1] 9
99
99
99
0
9
0
On Fri, Mar 6, 2020, 2:25 AM Henry Rich <[email protected]> wrote:
The current implementation is as you left it for this case.
>./vec times.
Henry Rich
On 3/5/2020 1:09 PM, Roger Hui wrote:
u is a verb and vec is a vector of non-negative integers. In u:vec 0, in
the current implementation, is u invoked +/vec times or >./vec times?
(Sorry if you are seeing this twice. Apparently my first post
disappeared.)
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