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10:50 PM
@Suever One of these days we need to assign & for r. Possibilities: (a) 2$ (b) 1$ (c) 1$ and make r with 1 input behave with an implicit leading 1, so 3&r would be rand([1,3]))
Yr with 1 input uses (c)
I think perhaps (c) would be best for r
 
11:06 PM
On the other hand, some answers (two) have used 1$r with the current meaning (one input specifies a square matrix). Maybe that's a useful meaning too
 
11:17 PM
@LuisMendo Yea I think that we'd probably want to go with the same as we use for l and O
So & == 2$r
I was pretty happy to use the do-twice loop though
 
11:53 PM
@Suever :-)
(a) If & is defined as 2$ (and no leading zero for single scalar input; that would be useless):
rand(1,3): l3&r
rand(3): 3t&r
rand(3,4): 3K&r
(b) If & is defined as 1$ (and no leading zero for single scalar input):
rand(1,3): l3h&r or l3H$r
rand(3): 3&r
rand(3,4): 3Kh&r or 3KH$r
(c) If & is defined as 1$ and leading 1 is implicitly added to single scalar input:
rand(1,3): 3&r
rand(3): 3th&r or 3t2$r
rand(3,4): 3Kh&r or 3K2$r
So its (a) 4,4,4 bytes versus (b) 5,3,5 versus (c) 3,5,5. Option (b) doesn't seem useful; the square case is not that common. Between (a) and (b), the question is, is the rand(1,n) case so much more common to warrant saving 1 byte at the cost of the other cases gaining 1 byte?
(a) is like l and O. (c) is like Yr
Hm no, Yr is different, because its default is 1$, not 0$
You are probably right: like l and O
 

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