[Rd] d-p-q-r-tests: why is plnorm(exp(Inf)) equal pnorm(Inf)?

[Thomas Hoffmann]

I stumbled across a result of d-p-q-r-tests:

d-p-q-r-tests expects pnorm(Inf) and pnorm(-Inf)  to give the same results as  plnorm(exp(Inf)) 
and plnorm(exp(-Inf)), respectively.

Does this mean that all expressions result in NaN (which I expect for the 3rd and 4th expression only) or 
does it mean that R can prevent the exp(Inf) to be evaluated before being parsed in the context of plnorm?

Thomas Hoffmann.

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