Examples of $I$-adically incomplete rings I seem to be short on examples for $I$-adic completions of rings.
I know that a ring is $I$-adically complete if the canonical homomorphism into the inverse limit is an isomorphism. My thinking and searching on the internet has been surprisingly fruitless, though, for examples where the map is either surjective and not injective, or injective and not surjective. (Am I mistaken that both are possible?)

So, the main question here is for one or more useful examples of both of these types of $I$-adically incomplete rings.

If possible, it would be nice to have the surjective-not-injective example be with respect to an ideal $I$ which contains a nonzero idempotent $e$.
 A: For injective but not surjective, take the ring $R = \Bbb{Z}$ and the ideal $I = \langle p \rangle$, for a prime $p$, to obtain the $p$-adic integers $\Bbb{Z}_{p}$. 
For surjective but not injective, I think you might take the ring $R = \Bbb{Z}_{p} \oplus B$, where $B$ is any Boolean ring, and $I = \langle (p, 0), \{0\} \oplus B \rangle$, the completion here being once more $\Bbb{Z}_{p}$, as $I^{n} = \langle (p^{n}, 0), \{0\} \oplus B \rangle$.
PS In a previous version I had $\lvert B \rvert = 1$. I expanded $B$ to any Boolean ring after rschwieb's comment below.
A: One thing I learned much later is: if $I$ is a nilpotent ideal, then $R$ is $I$-adically complete.
Thinking of the completion as a subring of $\prod R/I^n$, it's clear that at a certain $n$ the tail is just copies of $R$, and the congruence property that distinguishes elements of the product that are in the limit says that the tail of any given element is a constant sequence. Then it's not hard to see that the $r$ defining the constant tail is what maps onto your element.  So the canonical map is surjective. It's obviously injective too, so there's the isomorphism.
This provides a nice large class of $I$-adically complete rings. It's also worth noting that you can't weaken "nilpotent" to "nil" or even "T-nilpotent": there are counterexamples which aren't $I$-adically complete in those cases.
