In place behaviour of sbreak

[lcarver]

Hello,

I encountered this behaviour during the tutorial.

ring = at.load_mat('./dba.mat', mat_key='RING')

The following line does not do anything:
ring_tmp.sbreak(0.2, at.Marker('hello'))
and there is nothing here for copy=True or copy=False

at.sbreak does not exist so we cannot do:
at.sbreak(ring_tmp, 0.2, at.Marker('hello'))

Instead the working syntax was:
ring_tmp = ring_tmp.sbreak(0.2, at.Marker('hello'))

which seems a bit in contradiction to other lattice modification functions where the in_place modifications can be used.

[swhite2401]

sbreak modifies the lattice structure. An in place modification does not make much sense and therefore a new ring is returned.
sbreak is a member function of the class Lattice_object so it is only accessible through it and in my opinion this should remains like this.

The only improvement I can see would be to put an example usage in the help. Would that be suitable for you?

[lcarver]

Yes I think it would make it clearer. I am happy to do the PR (within the next few days)