Thanks for the diagram, but it does not answer my point, let alone refute it. Of the two sentences you quoted, you only attempt to respond to one as far as I can tell.

Saying "the context of the assignment operator is scalar, which changes what particular representation of the list is propagated" does nothing to show that there's no context applied to the list. Saying the values propagate from right to left through the contexts is equivalent to saying that the contexts propagate from left to right through the values.

Now, about that second sentence you quoted... If there is no list, how can there be a list assignment? In what language are you writing that looks so much like English but in which "list assignment" does not mean "assignment of a list"?

What does "So while the list assignment is in scalar context, the assignee and assigned lists aren't." if the "list" doesn't exist?

What the heck is "list op'tor"? Does that stand for "list operator"? Upon what does a "list operator" operate?

Show me the list assignment that is in scalar context here: 

$ perl -E 'say scalar(() = ($a, $b, $c))'
               |      |
                \      \_____ is a list?
                 \
                  \__________ enforces scalar context?

                     ?_______ second list assignment?

[download]

Is there a list there? Is there a set of scalar values context assignment operator present other than the one between the empty set of scalar values and the set of three scalar values? In what context is the empty set of scalar values?

I don't believe you've shown there's no concept of a list being coerced into a scalar context. I especially don't believe you've shown there's no concept of a list in Perl.

In reply to Re^10: If you believe in Lists in Scalar Context, Clap your Hands by mr_mischief
in thread If you believe in Lists in Scalar Context, Clap your Hands by gone2015

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