On 17.06.2011, at 2:02AM, Mark Wiebe wrote:
>> ok, that was a lengthy hunt, but it's in printing the string in
>> make_iso_8601_date:
>>
>> tmplen = snprintf(substr, sublen, "%04" NPY_INT64_FMT, dts->year);
>> fprintf(stderr, "printed %d[%d]: dts->year=%lld: %s\n", tmplen, sublen,
>> dts->year, substr);
>>
>> produces
>>
>> >>> np.datetime64('1970-03-23 20:00:00Z', 'D')
>> printed 4[62]: dts->year=1970: 0000
>> numpy.datetime64('0000-03-23','D')
>>
>> It seems snprintf is not using the correct format for INT64 (as I happened
>> to do in fprintf before
>> realising I had to use "%lld" ;-) - could it be this is a general issue,
>> which just does not show up
>> on little-endian machines because they happen to pass the right half of the
>> int64 to printf?
>> BTW, how is this supposed to be handled (in 4 digits) if the year is indeed
>> beyond the 32bit range
>> (i.e. >~ 0.3 Hubble times...)? Just wondering if one could simply cast it to
>> int32 before print.
>>
> I'd prefer to fix the NPY_INT64_FMT macro. There's no point in having it if
> it doesn't work... What is NumPy setting it to for that platform?
>
Of course (just felt somewhat lost among all the #defines). It clearly seems to
be mis-constructed
on PowerPC 32:
NPY_SIZEOF_LONG is 4, thus NPY_INT64_FMT is set to NPY_LONGLONG_FMT - "Ld",
but this does not seem to handle int64 on big-endian Macs - explicitly printing
"%Ld", dts->year
also produces 0.
Changing the snprintf format to "%04" "lld" produces the correct output, so if
nothing else
avails, I suggest to put something like
# elseif (defined(__ppc__) || defined(__ppc64__))
#define LONGLONG_FMT "lld"
#define ULONGLONG_FMT "llu"
# else
into npy_common.h (or possibly simply "defined(__APPLE__)", since %lld seems to
work on 32bit i386 Macs just as well).
Cheers,
Derek
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