I'm just going through perlxstut and I found there newSVnv in EXAMPLE 5 and EXAMPLE 6 but I think that newSVuv should be more appropriate. Curiously newSVnv works too. What's going on?
+2
A:
I think it uses NVs (Perl's equivalent of a C double) instead of UVs (normally an unsigned int), because (depending on OS and compilation options), some of the values in a struct statfs might be 64-bit even though Perl is using 32-bit ints. newSVnv works because the C compiler knows how to cast any integer type to a double.
You should be able to replace newSVnv with newSVuv for any member of statfs that will fit in a UV, and have it work just fine. Perl converts between its numeric types automatically as needed.
cjm
2010-09-28 21:36:21
So it means that `newSVnv` is considered safer version of `newSVuv` in Perl sense for Perl's automatic conversion?
Hynek -Pichi- Vychodil
2010-09-28 23:07:53
I'm not sure what you mean by "safer". There's no automatic conversion with the `newSV*v` functions (other than that done by the C compiler). If you pass a `long long` to `newSVuv`, then you get whatever happens when your C compiler casts a `long long` to `unsigned int`. A `double` has a greater range than any integer type, so you don't have to worry about a value that won't fit.
cjm
2010-09-28 23:38:17
@cjm: that is incorrect; many systems have 8 byte doubles (with 53 bites of precision) and 8 byte integers. storing them in a double can lose precision.
ysth
2010-09-29 02:05:06
@ysth, that's why I said "range" and not "precision". You might lose precision when converting a large integer type to a `double`, but it won't be out of range.
cjm
2010-09-29 02:21:30