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Re: unexpected result of stringification.

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Chris Stith
March 19, 2001 13:53
Re: unexpected result of stringification.
Message ID:
On Mon, 19 Mar 2001 wrote:

> Chris Stith <> writes:
> >That is so if the examples are taken exactly as is. It is not so if
> >the example is put in the context of code following it. From which
> >match operator would the next line pull `$1'?
> >
> >Nick is talking about a body of code expecting something to be a
> >certain way when the code example above is executed. 
> No I am not. (The example code is not mine and not a good example.
> It has && in it, is calling a builtin as a pseudo-sub and ...)
> I am talking about code that uses $1 in general, and with at least
> a basic grasp of how the op tree is built. And with a background

> in devious "lexing" schemes. 

 if( underlined_section_is_flame ) {
    Thank you for assuming that because I'm not as intimately familiar
    with perl or with your own explanation of this issue as are you
    that I have no familiarity with lexing or parsing.

The thread, as quoted by Abigail, read to me as if you meant that
an existing body of code expected one or the other match on a single
line to become `$1'. I realized it was not your example, but the
context in which I was reading I understood to be proper context
for what you said. I must have misunderstood what point Abigail
was trying to make.

> Without a lot of flow analysis (that perl does not do) perl cannot tell 
> that a particular use of $1 is (always) used in a way that means it should
> take a copy, or a way which means it should leave it "lazy".
> So my stance is that $1 is lazy, if you want/need a copy then write "$1".

That's something on which I am not prepared to comment, given that I
haven't studied that section of the interpreter thoroughly enough. The
only thing I will say is that if there's a way to do either the
programmer wishes, which you seem to say there is, then that should
be a definite plus compared to forcing `$1' itself one way or the other.
If `$1' is lazy now and can be forced to be a copy when wanted, then
that makes for the best flexibility.

> >A line of
> >code following the above example shouldn't expect one or the other
> >expression to be either `first' or `last', let alone expect which one
> >will be the value of `$1'.
> Things get less clear when you have something like:
>   some_sub( (s/(0x([a-f0-9]+))/another_sub($1)/ie || s/(\d+)/third($1)/e) ?  $1 : ... );
> In such cases "lazy" is what _I_ want for that last one at least and only 
> by a lot of looking do you know which that is.

I notice you underline `I'. That's a good habit in a discussion such
as this. ;-)

Chris Stith

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