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Understanding the more common-place notion of right-associativity is not enough.
Indeed. You must also know the return value of the operator (= in this case). But that's the same as trying to understand the result of
2 ** 3 ** 4
you need to understand the return value of ** beside understanding right-associativity.
I do not know how this is actually implemented.
Associativity is a parser/compiler issue. By the time the code is running, associativity no longer plays a role. As for the implementation of assignment, they are actually two different operators, scalar assignment (pp_sassign in pp_hot.c) and list assignment (pp_aassign in pp_hot.c).

And if you look in pp_hot.c, you see there's check on context to determine what is returned:

if (gimme == G_VOID) SP = firstrelem - 1; else if (gimme == G_SCALAR) { dTARGET; SP = firstrelem; SETi(lastrelem - firstrelem + 1 - duplicates); } else { if (ary) SP = lastrelem; else if (hash) { if (duplicates) { /* Removes from the stack the entries which ended up a +s * duplicated keys in the hash (fix for [perl #24380]) + */ Move(firsthashrelem + duplicates, firsthashrelem, duplicates, SV**); lastrelem -= duplicates; } SP = lastrelem; } else SP = firstrelem + (lastlelem - firstlelem); lelem = firstlelem + (relem - firstrelem); while (relem <= SP) *relem++ = (lelem <= lastlelem) ? *lelem++ : &PL_sv_undef; } RETURN;
Now, I don't expect you to understand everything (neither do I), but I do hope you see that in VOID context, nothing is returned, in SCALAR context, a single value is returned, and in LIST context, more things are returned.

One more piece of evidence that it isn't the scalar assignment that is reaching all the way behind, here we have a list assignment in scalar context where the scalar context isn't given by another assignment:

$ perl -E 'say scalar(() = ($a, $b, $c))' 3 $

In reply to Re^7: If you believe in Lists in Scalar Context, Clap your Hands by JavaFan
in thread If you believe in Lists in Scalar Context, Clap your Hands by oshalla

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