there aren't really any surprises in the actual BF interpretation.

## but I must admit that I needed a -MO=Deparse to get to that 's<>><\
+>>e'
## thingy. I still don't understand how can it be equivalent to
## 's//<ARGV>;/e'.
[download]

s<>><\>>e is a s///e where the first pair of delimiters is <>, and the second pair is >>. The replacement eval is <>, because the \ in <\> gets removed earlier as it's right before the same character as the delimiter. It's the same form as s()/<>/e.