Why is implicit and explicit 'nil' treated differently?

As you can see in the sample code below, print does not show the implicit nil (when it is last in the list of things to print), yet it shows the explicit one. However, in every other regard, nil seems to behave the same in both cases.

Can someone explain why there is a difference in 'nil' treatment between these two?

(I can see the compiler produces slightly different code for each case but if they are equivalent, shouldn’t it produce the exact same code for both cases?)

And, is the difference evident only in print or elsewhere also?

I’m mostly interested if I should expect the exact same execution path for code receiving either nil, or not? Because obviously print has a different execution for each case.

Thank you.

function implicit()

return --implicit nil

end

function explicit()

return nil --explicit nil

end

print(implicit()==nil,implicit() or 'hidden',implicit())

print(explicit()==nil,explicit() or 'shown',explicit())

print(implicit() == explicit())

--- And here’s what the compiler output looks like for the two functions ---

function <implicit> (2 instructions at 004CC030)
0 params, 2 slots, 0 upvalues, 0 locals, 0 constants, 0 functions
1 [2] RETURN 0 1
2 [3] RETURN 0 1

function <explicit> (3 instructions at 004CE790)
0 params, 2 slots, 0 upvalues, 0 locals, 0 constants, 0 functions
        1       [6]     LOADNIL         0 0
        2       [6]     RETURN          0 2
        3       [7]     RETURN          0 1