Re: Breaking a table of record rows into an array

Kaz Kylheku <433-929-6894@kylheku.com> writes:

On 2024-03-13, Keith Thompson <Keith.S.Thompson+u@gmail.com> wrote:

Kaz Kylheku <433-929-6894@kylheku.com> writes:

On 2024-03-13, Keith Thompson <Keith.S.Thompson+u@gmail.com> wrote:

arnold@freefriends.org (Aharon Robbins) writes:

In article <usqkgn$he7u$2@dont-email.me>,
Ed Morton  <mortonspam@gmail.com> wrote:

the effect of setting `NF` is
undefined behavior per POSIX and so will do different things in
different awk variants and even in 1 awk variant can behave differently
depending on whether you're setting it to a higher or lower than
original value

>
This is not true. The effect of setting NF was well defined
by the original awk book and also in POSIX.
>
Decreasing NF throws away fields. Increasing NF adds the
intervening fields with the null string as their values
and rebuilds the record.

>
I don't see that in the POSIX specification.

>
The key is this:
>
  References to nonexistent fields (that is, fields after $NF), shall
  evaluate to the uninitialized value.
>
NF is assignable, and fields after $NF do not exist. Thus if we
have four fields and set NF = 3, then $4 doesn't exist.

>
That describes what happens if NF is modified by assignment, but I don't
see that it implies that such an assignment is allowed.

>
"The left-hand side of an assignment and the target of increment and
decrement operators can be one of a variable, an array with index, or a
field selector."
>
NF is described as a variable.  Some unique remarks are made about NF,
but none deny that it's assignable like any other variable.

OK, I concede.  It can be inferred from the POSIX specification that
assigning to NF is allowed.

And the specification is in serious need of a definition of what
assigning to NF actually *does*, other than changing the value of NF.

But I can imagine a hypothetical awk-like language in which assigning to
NF has undefined behavior.  My question is, how does the POSIX
specification not describe that language?

>
That language is failing to support an instance of a variable
being the left operand of an assignment, which a variable "can be".
>
It looks like the violation of a requirement.

Agreed.  I think.

[...]

--
Keith Thompson (The_Other_Keith) Keith.S.Thompson+u@gmail.com
Working, but not speaking, for Medtronic
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