I have a quick question:
if a=1, h=1, b=5, then a:h:b will turn out to be [1 2 3 4 5]
However, if I want to do backward: b:h:a
it turns out to be a 1*0 empty row vector
is there any possibility we can still equally distribute the space and make b:h:a turns out to be [5 4 3 2 1]?

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KSSV
KSSV am 10 Nov. 2021

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a=1;
h=1;
b=5 ;
l = a:h:b
l = 1×5
1 2 3 4 5
m = b:-h:a
m = 1×5
5 4 3 2 1

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