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Re^8: If you believe in Lists in Scalar Context, Clap your Hands

by mr_mischief (Monsignor)
on Oct 24, 2008 at 17:51 UTC ( #719394=note: print w/ replies, xml ) Need Help??


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

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:

If you can't have a list in a scalar context, how can you have the assignment of a list in scalar context? You have even said in this thread that lists don't even exist.

To pick your own nit, perhaps you meant "the result of the list assignment operator upon assigning a series of scalars to an empty series of scalars".

It's very difficult for there to be a list assignment in scalar context there if lists don't exist. What does a "list assignment" assign? It's also very difficult to support the idea that lists don't exist when you show the code that returns multiple scalars ("atoms" and references to other complex structures) in order (a "list") in LIST context with variables named 'lelem' and similar. The "elem" part of those clearly stands for "element". Elements of what?

You can argue that there is another, clearer concept to use to understand this topic. It's very difficult to support the idea that the concept used in the documentation, the source code, the tutorials, and the in concepts and culture of other programming languages from which Perl borrows ideas does not exist in Perl, though. Call it flawed if you like, but saying it doesn't exist is inaccurate.


Comment on Re^8: If you believe in Lists in Scalar Context, Clap your Hands
Re^9: If you believe in Lists in Scalar Context, Clap your Hands
by ikegami (Pope) on Oct 24, 2008 at 20:03 UTC

    If you can't have a list in a scalar context, how can you have the assignment of a list in scalar context? You have even said in this thread that lists don't even exist.

    The list assignment imposes a list context on its operands. So while the list assignment is in scalar context, the assignee and assigned lists aren't.

    A B C D <-- labels $x = ( $i, $j ) = ( $p, $q ); S ? L L L S L L L <-- context | | | | | | | | \__ Propogated by right-most list op'tor (D). | | | | | | | \____ Imposed by the list assignment (C). | | | | | | \______ Propogated by right-most list op'tor (D). | | | | | \__________ Imposed by the scalar assignment (A). | | | | \_______________ Propogated by left-most list op'tor (B). | | | \__________________ Imposed by the list assignment (C). | | \___________________ Propogated by left-most list op'tor (B). | \_______________________ Determined by outside code. \__________________________ Imposed by scalar assignment (A).

      Now try and replace all the information conveyed by the use of the word 'list' in the following sentence from perlsub, with the one or two dozen diagrams like the above, it would take to replace that one word.

      If you return one or more aggregates (arrays and hashes), these will be flattened together into one large indistinguishable list.

      The only purpose served by the "there's no such thing as a list in a scalar context" meme, is the promotion of one person's intimate knowledge of perl's internals over another's lack thereof; or lack of interest in same.


      Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
      "Science is about questioning the status quo. Questioning authority".
      In the absence of evidence, opinion is indistinguishable from prejudice.

        The quote is true. Of course, the only way to return more than one aggregate is to be in list context.

        If a flattened list was being returned in the following code, it would print "d" instead of "4".

        perl -le"print scalar sub { my @a = qw(a b c d); return (@a,@a) }->(); +"

        The only purpose served by the "there's no such thing as a list in a scalar context" meme

        So? I'm against it too. Too many definitions of list involved. I've been promoting "an operator/function/sub/expression can't return a list in scalar context".

      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?

      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.

        Of the two sentences you quoted, you only attempt to respond to one as far as I can tell.

        The inclusion of the second sentence was accidental. I hadn't even read it. If JavaFan said that lists don't even exist, that I don't agree with him. I suspect that's not what he meant, though. I can't address this question for him.

        Saying [...] does nothing to show that there's no context applied to the list.

        I agree. Nothing will show that because everything is evaluated in some context.

        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.

        Operands are evaluated in a context chosen by their operator. Precedence and associativity determine what is an operand to what operator, but doesn't otherwise affect context.

        To clarify, by "an operator propagating a context onto an operand", I mean "an operator choosing to impose the same context upon an operand as the context imposed upon the operator itself".

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

        Where did you get that from? I didn't say some list didn't exist. Quite the opposite, I showed two list operators in list context, so two lists were created.

        What the heck is "list op'tor"? Does that stand for "list operator"?

        Yup. I wanted to avoid reaching the default wrap length for code blocks.

        () -- is a list?

        Yes, an empty list literal.

        scalar -- enforces scalar context?

        Yes, onto its operand, the list assignement.

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

        | v perl -E 'say scalar(() = ($a, $b, $c))' V L L S L \ L/ L \/ L

        Is there a list there?

        Multiple.

        • () (or rather the nothing inside the parens) is a list operator.
        • ,, is a list operator (or two depending if you think of it as a binary operator).

        • () returns a list of zero scalars.
        • ($a, $b, $c) returns a list of three scalars.
        • scalar(...) returns a list of one scalar.

        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?

        What's a "scalar values context assignment operator"?

        In what context is the empty set of scalar values?

        You mean the ()? List context. The () is the LHS of a list assignment operator, and a list assignment operator imposes a list context on its LHS operand. (On its RHS operand too.)

Re^9: If you believe in Lists in Scalar Context, Clap your Hands
by JavaFan (Canon) on Oct 25, 2008 at 22:52 UTC
    If you can't have a list in a scalar context, how can you have the assignment of a list in scalar context?
    Because any operator can be scalar context; it's the context that determines its context, not their operands. Here are a few examples of operators in scalar context:
    scalar ($a + $b); # Addition scalar (($a) x $b); # Repetition scalar ($a =~ $b); # Matching scalar ($a .. $b); # Range/FlipFlop scalar ($a = $b); # Scalar assignment scalar (($a) = ($b)); # List assignment
    Here are some examples of operators in list context:
    say ($a + $b); # Addition say (($a) x $b); # Repetition say ($a =~ $b); # Matching say ($a .. $b); # Range/Flipflop say ($a = $b); # Scalar assignment say (($a) = ($b)); # List assignment
    One should realise that 'scalar assignment' and 'list assignment' are two different operators (sassign and aassign in pp_hot.c). It's the LHS of the '=' that determines which operator it is (this is unlike the '..' operator where it's the context that determines whether it's the range or the flipflop operator). The side-effects (that is, the actual assignment) of both sassign and aassign are independent of the context. However, the return value of the aassign operator is context dependent. In scalar context, it returns the number of elements appearing on its RHS.

      it's the context that determines its context

      Deep recursion on "context" at PerlMonks post 719579 Terminating on signal SIGINT(2)
      :)
      Because any operator can be scalar context, including the list operator.
        How does the "list operator" look like?
Re^9: If you believe in Lists in Scalar Context, Clap your Hands
by ysth (Canon) on Oct 28, 2008 at 06:27 UTC
      No. Just, no.

      People are saying there are only three little things to remember so their model is much simpler.

      The ideas that "list" is a "value", "assignment" is an "operation", and that a "list of scalars" in the syntax does not become a "list value" in the semantics are at least three more concepts to that abstraction.

      It's not a non-question, because I can answer how it happens: a list of scalars doesn't propagate its own value, and in an assignment is processed by the assignment operator which in a scalar context returns the last value in the list. This last value is typically the last value returned by the comma operator.

      Without the additional concepts of "list of scalars" vs. just "list", "list value" vs. just "list", the idea that the syntactic "list of scalars" is not what propagates the value, that the assignment operator does, and just what the comma operator is doing (or that it is an operator at all and not just part of a "list syntax"), the simple "three" rule explanation doesn't explain anything to the uninitiated.

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