Overloading relational operators separately; thoughts?

[Chris Wright via Digitalmars-d]

On Thu, 29 Sep 2016 17:50:54 -0700, Jonathan M Davis via Digitalmars-d 
wrote:
> Except that it kind of is. It's an example of a language allowing you to
> mess with too much and make it so that it doesn't function as expected,
> which is what happens when you overload operators to act in a way
> inconsistent with how they work with the built-in types.

The perl example is a line of code buried somewhere that changes the 
meaning of the rest of the code. Operator overloading is restricted to a 
specific user-defined type. With such a dramatic difference in the scope 
of the change, the analogy is useless.