Steven D'Aprano added the comment:
Why would they give the same result when the code is different?
#1 assign c, d, e and everything left over goes into b
*b, c, d, e, = [1, 2, 3, 4]
#2 assign c, d and everything left over goes into b
*b, c, d, = [1, 2, 3, 4]
#3 assign c a
New submission from Michael Jaquier :
x = [1,2,3]
Case (1)
*b, = x
*b == [1, 2, 3]
Case(2)
*b, _ = x
b = [1,2]
Is it not more logical for this to give consistent values regardless.
i.e.,
*b, = [1,2,3] ; b == [1,2]
*b, _ = [1,3,2] ; b == [1,2] ; _ == [3]
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components: Libra
