in reply to Re^4: What's most efficient way to get list context? (count)
in thread What's most efficient way to get list context?

when perl is coaxed into evaluating the RHS in a list context, and then the returned list is put in a scalar context,
No, you can't "put {a list} in a scalar context".

What you've done is demonstrate that the list-assignment operator ((...) = LIST) returns the number of elements assigned in a scalar context. This is both documented, and could just as easily invoke Nethack instead (although that would be less useful).

That has nothing to do with "list in scalar context", because "list in scalar context" can't exist.

It helps me to always think of the abstract syntax tree. Every node in the tree is some kind of operator/function. Every node is being evaluated in some sort of larger context (scalar/list/void, and to a lesser extent scalar can also be numeric or string). Every node can choose to return different things to each of those contexts.

In your example, the "top node" is "list assignment operator". If you evaluate that in a list context, it returns a copy of the contents. If you evaluate that in a scalar context, you get the number of elements copied.

At no time is there a "node" that has a strictly list value that is being evaluated in a larger context. In some sense, lists don't exist, except because someone applied list context to some node that can return a scalar or a list. Because there's no existing node that returns a list in a scalar context, it simply does not exist. You could probably write one in XS, and break things, but there's no existing one that does that. {grin}

-- Randal L. Schwartz, Perl hacker
Be sure to read my standard disclaimer if this is a reply.

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Re^6: What's most efficient way to get list context? (count)
by Roy Johnson (Monsignor) on Apr 15, 2005 at 15:18 UTC
    I submit that a slice is a form of a list. By "a form of a list", I mean that its behavior is exactly like the literal list of the elements would be. In scalar context, it returns its last element, but there is no comma operator at work. Is that undocumented behavior?

    I think it's reasonable behavior, and I think it's reasonable to have the concept of a list apart from the comma operator.

    Caution: Contents may have been coded under pressure.
[reply]
      A slice is an operator, yet another kind of "node" using my terminology up-thread. A slice in a list context does the listy thing. A slice in a scalar context returns the last element, but could have been made to invoke nethack again. It has nothing to do with the comma operator, which has its own scalar/list behavior.

      The only consistency is that Larry made some of them consistent. It's DWLM: "Do What Larry Means". But it is learnable, and does make sense once you learn it.

      -- Randal L. Schwartz, Perl hacker
      Be sure to read my standard disclaimer if this is a reply.

[reply]
        A slice in a scalar context returns the last element, but could have been made to invoke nethack again.
        Yes, it could have, but it wasn't, and so the concept of multi-context LISTing is reinforced. My point is that There Is More Than One Way To Think About It, and unless there is some behavior that breaks the mental model, there's nothing wrong with the concept of a list in a scalar context. It explains WLM more simply than a whole bunch of special cases that we pretend are absolutely unrelated. They are related by Larry's vision, which appears to include the notion of a LIST.

        The comma is a LIST-construction operator that works in either context, and the slice is another. qw is another. The range operator is not, because in scalar context, it's a completely different operator.

        Caution: Contents may have been coded under pressure.
[reply]

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