Re^2: A little overloading conundrum

There's nothing I can think of and the documentation suggests it's not something that is easy to do.

Yes - the first point I read in that link to the documentation pretty much kills all hope:

1.If the first operand has declared a subroutine to overload the opera
+tor then use that implementation.
[download]

I suppose (untested) I could do:

my $n2 = (\$B_obj) - $A_obj;
[download]

and that would at least call module A's oload_minus() subroutine .... which would then be structured to de-reference the first argument and return the intended result.
But that solution is no more practical than the alternative you provided.

Thanks for the reply. I had, of course, consulted the overloading docs but had stopped reading before reaching the bit to which you linked.
It's not the end of the world if I have to modify module B.

Cheers,
Rob

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Re^3: A little overloading conundrum by ysth (Canon) on Mar 08, 2026 at 03:22 UTC
If you are working around it in the caller, this is clearer, if not as efficient: `my $n2 = -($A_obj - $B_obj);` [download]	[reply] [d/l]
Re^4: A little overloading conundrum by syphilis (Archbishop) on Mar 08, 2026 at 04:38 UTC
If you are working around it in the caller... Apologies - I should have explicitly stated that "working around it in the caller" is not a solution. It's imperative that, wrt the given pseudo-example, `my $n2 = $B_obj - $A_obj;` return a module A object with a value of -10. Hence my agreement with sw1's labelling of both his and my alterations to the caller as being "not practical". If module B did not overload the '-' operator then, AIUI, `my $n2 = $B_obj - $A_obj;` would invoke module A's overloading of that operator - which is exactly what I want. I was hoping there might have been some way of triggering that same behaviour when module B does overload that operator .... but it seems that's not the case. Cheers, Rob	[reply] [d/l] [select]