Why would a branch cut not end at a branch point?

Both Wikipedia and MathWorld (here and here) seem to place some imporantance on saying, but not elaborating on

It should be noted that the endpoints of branch cuts are not necessarily branch points.

When would it make sense to have a branch cut that ends at a non-branch point?

Is it just because one can concoct spurious cases such as $$ f(z) = \begin{cases} 1 & |z| < 1 \\ 2i & |\Im(z)| > 2 \\ z & \text{otherwise} \end{cases}$$ with a "branch" cut with no ends along the unit circle, and another one going from infinity to infinity? Or is there more to it?

Hmm, here is one arguable case:

The lacunary function $f(z)=\sum_{n=0}^{\infty} z^{n!} $ is defined on the unit disc and has singularities at every point on the unit circle. Now, the function $$ g(z) = f(e^{-\sqrt z}) $$ where the square root is the usual principal square root, is defined on the complex plane except for the (closed) negative real axis.

Then one might consider the negative real axis to be a branch cut for $g$. Or perhaps not; that depends on exactly what one takes "branch cut" to mean and how seriously one takes the requirement (in both Wikipedia's and Wolfram's definitions) that the function must be "multi-valued".

In any case its endpoint at $0$ is definitely not a branch point: one cannot continue the function analytically along any closed curve around $0$.

Alternatively, and less dramatically, one may consider $$ h(z) = \sum_{n=1}^\infty \frac{\sqrt{z+1/n}}{2^{-n}} $$ where again all of the square roots are principal ones. Here the discontinuity along the negative real axis is actually a bona fide branch cut at all except countably many points -- but still the endpoint at $0$ is not a branch point.