This still doesn't answer the question as to why foo2() receives as pre-existing scalar makes no changes to it and returns it to the caller, and suddenly we have an unreferenced scalar message.

Which scalar suddenly becomes mortal just by virtue of being passed into a subroutine and returned? As there is only one scalar involved in foo2(), it can only be SV *a, but that brings me back to the question of why?

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
Lingua non convalesco, consenesco et abolesco. -- Rule 1 has a caveat! -- Who broke the cabal?
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.

In reply to Re^2: XS/Inline::C concat *any* two SVs. by BrowserUk
in thread XS/Inline::C concat *any* two SVs. by BrowserUk

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