you could start with a couple of IV's and generate a NaN.

The problem (for me) is that there are no IVs involved in your construction.

The basic compile-time constant: 2**1025 never results in the construction of any IVs.

The interpreter constructs a compile-time constant from the expression and stores it as an NV:

Perl> use Devel::Peek;;
Perl> $x = 2**1025;;
Perl> Dump( $x );;
SV = NV(0xb4598) at 0x3d73ee8
  REFCNT = 1
  FLAGS = (NOK,pNOK)
  NV = 1.#INF
[download]

And then at runtime, one NV (value 1.#INF) is divided by another NV (also 1.#INF) and the results is an NV (value: -1.#IND):

$y = $x / $x;;
Dump $y;;
SV = NV(0xb45a0) at 0x3d6c6a8
  REFCNT = 1
  FLAGS = (NOK,pNOK)
  NV = -1.#IND
[download]

No IVs were ever involved because the textual integer expression from the source code never made into the byte-coded Perl.

The problem (for me) is that there are no IVs involved in your construction



C:\>perl -MDevel::Peek -le "Dump(2);"
SV = IV(0x1daf1b8) at 0x1daf1bc
  REFCNT = 1
  FLAGS = (PADTMP,IOK,READONLY,pIOK)
  IV = 2

C:\>perl -MDevel::Peek -le "Dump(1025);"
SV = IV(0x53ee70) at 0x53ee74
  REFCNT = 1
  FLAGS = (PADTMP,IOK,READONLY,pIOK)
  IV = 1025
[download]

Or, alternatively, I could have assigned the values 2 and 1025 to two IV's.

The interpreter constructs a compile-time constant from the expression and stores it as an NV.

Yes, nothing remarkable about that - most perl users (you and I included) are aware of this and generally comfortable with it.
However, it implies to me that in perl, there are no distinct realms of ints and floats. And your remark that "NaN is a purely floating point concept" (which is correct) suddenly seemed a bit blurry wrt perl - because, in so many ways, perl doesn't draw a line between floats and ints - yet it does have NaNs.

Anyway ... I should have tried harder to explain the triviality or, better still, just made a mental note of it and said nothing.

I sure am glad I didn't create a demo where the upgrade from IV to NV occurred because of integer overflow ... then manipulate the NV to create a NaN ... then claim to have shown that a NaN *can* arise from integer overflow. Something like:

C:\>perl -le "$x= ~0; $x += 2; $x **= 35; $x /= $x; print $x;"
-1.#IND
[download]

;-))

(Complete and utter sophistry, of course.)

Cheers,
Rob

Hm. $x starts undefined

[0] Perl> use Devel::Peek;;
[0] Perl> Dump $x;;
SV = NULL(0x0) at 0x3c6d400
  REFCNT = 1
  FLAGS = ()
[download]

You set $x to the maximum unsigned integer value; it becomes an IV:

[0] Perl> $x = ~0;;
[0] Perl> Dump $x;;
SV = IV(0x3c6d3f8) at 0x3c6d400
  REFCNT = 1
  FLAGS = (IOK,pIOK,IsUV)
  UV = 18446744073709551615
[download]

You add 2 to it, it overflows the IV, so perl converts it to an NV (NB: No NaN arises from the overflow of the IV.)

[0] Perl> $x += 2;;
[0] Perl> Dump $x;;
SV = PVNV(0x3cb85e8) at 0x3c6d400
  REFCNT = 1
  FLAGS = (NOK,pNOK)
  IV = -1
  NV = 1.84467440737096e+019
  PV = 0
[download]

Then, you raise that NV to the power 35; and the NV overflows; resulting in an NV with the floating point value of positive infinity (1.#INF) (NB:the "result to large" error):

[0] Perl> $x **=35;;
[Result too large] Perl> Dump $x;;
SV = PVNV(0x3cb85e8) at 0x3c6d400
  REFCNT = 1
  FLAGS = (NOK,pNOK)
  IV = -1
  NV = 1.#INF
  PV = 0
[download]

You then divide the NV 1.#INF, by itself, and you produce an indeterminate floating point value (1.#IND):

[0] Perl> $x /= $x;;
[0] Perl> Dump $x;;
SV = PVNV(0x3cb85e8) at 0x3c6d400
  REFCNT = 1
  FLAGS = (NOK,pNOK)
  IV = -1
  NV = -1.#IND
  PV = 0
[download]

The (special form of) NaN was produced by operating upon (erroneous) floating point (NV) values; not integer (IV) overflow.

NaN is a purely floating point concept, and cannot arise as a result of integer overflow.

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