Conventionally your size_type should be the same type returned by size() and used for indexing. So I would expect that type to be int, given the above.
Having said that, you're already departing from standard container conventions by having size() return a number that is *sometimes* 2 smaller than the "real" number of bins, which might frustrate generic algorithms.
Completely without tongue in cheek, I wonder if it might be better to not provide a size() method at all (to avoid container implications which you do not satisfy) and spell it as bin_count() or something distinct instead.
Another possibility that might sidestep all of this could be to remove the underflow/overflow bins from the standard indexing and make them only accessible through separate underflow_bin() methods (or similar) instead. But this might complicate other cases such as a find_bin_for_value() that needs to return an underflow/overflow (though that could be solved by returning a bin iterator, not an index).
A high +1:
Having size() makes on expect that "for(i=0;i I think representing the underflow bin with -1 and the overflow bin
with the value returned by size() is very intuitive and elegant. IMO this is against expectation of every C++ programmer: Being able to
index a container with -1 and size() so in every review of code using
this it will come up at least as a raised eyebrow.
Other idea: If those bins are so special that they don't fit into the
[0, size()) range, why not use a different function for getting them,
which is not the index operator? high_bin()/low_bin() come to mind.
But WHY was this chosen? Wouldn't it be ok if 0 is the first bin which
starts at -inf and size()-1 to be the last one spanning to inf? This
would allow a histogram of size 1 which has a single bin holding all
values. This makes me wonder to if you *always* want the +-inf bins or
if there are reasonable use cases NOT wanting them and expect
out-of-range values to be dropped.
Disclaimer: I'm not familiar with Boost.histogram and see this from a
users perspective only.