Hi,
UPDATE: I've now filed a
bug report about this.
I've been chewing on this over the last couple of days, and it doesn't look right to me.
It's an issue that arises only when NVprecision is greater than IVprecision.
So, if ivsize is 8 bytes, then in order to experience the issue, you'll generally need to use a quadmath build (nvtype of __float28, NVprecision of 113 bits).
With such a perl, I get:
C:\_32>perl wle "$m = (2**113)  1; $n = 2; print $m % $n;"
0
In contrast, POSIX::fmod returns "1" for the same operation.
C:\_32>perl MPOSIX wle "$m = (2**113)  1; $n = 2; print fmod($m, $n
+);"
1
Here's the section of perlop documentation that I'm looking at:
Binary "%" is the modulo operator, which computes the division remaind
+er
of its first argument with respect to its second argument. Given integ
+er
operands $m and $n: If $n is positive, then "$m % $n" is $m minus the
largest multiple of $n less than or equal to $m. If $n is negative, th
+en
"$m % $n" is $m minus the smallest multiple of $n that is not less tha
+n
$m (that is, the result will be less than or equal to zero). If the
operands $m and $n are floating point values and the absolute value of
$n (that is "abs($n)") is less than "(UV_MAX + 1)", only the integer
portion of $m and $n will be used in the operation (Note: here "UV_MAX
+"
means the maximum of the unsigned integer type). If the absolute value
of the right operand ("abs($n)") is greater than or equal to
"(UV_MAX + 1)", "%" computes the floatingpoint remainder $r in the
equation "($r = $m  $i*$n)" where $i is a certain integer that makes
+$r
have the same sign as the right operand $n (not as the left operand $m
like C function "fmod()") and the absolute value less than that of $n.
Note that when "use integer" is in scope, "%" gives you direct access
+to
the modulo operator as implemented by your C compiler. This operator i
+s
not as well defined for negative operands, but it will execute faster.
AFAICT, the parts of that documentation that are pertinent to my given example are:
If $n is positive, then "$m % $n" is $m minus the largest multiple of
+$n less than or equal to $m.
....
If the operands $m and $n are floating point values and the absolute v
+alue of
$n (that is "abs($n)") is less than "(UV_MAX + 1)", only the integer
portion of $m and $n will be used in the operation (Note: here "UV_MAX
+"
means the maximum of the unsigned integer type)
And to me that implies that the correct calculation for the given example is to do:
C:\_32>perl wle "$r = 10384593717069655257060992658440191.0  (519229
+6858534827628530496329220095.0 * 2); print $r;"
1
This agrees with the value produced by POSIX::fmod(), but disagrees with the value produced by the "%" operator.
Do we agree that perl is buggy here ?
If you have a perl with ivsize of 4, you can also demonstrate the same issue.
With ivsize of 4, NVprecision of 113 bits:
C:\_32>perl wle "$m = (2**113)  1; $n = 2; print $m % $n;"
0
C:\_32>perl MPOSIX wle "$m = (2**113)  1; $n = 2; print fmod($m, $n
+);"
1
With ivsize of 4, NVprecision of 64 bits (long double):
C:\>perl wle "$m = (2**64)  1; $n = 2; print $m % $n;"
0
C:\>perl MPOSIX wle "$m = (2**64)  1; $n = 2; print fmod($m, $n);"
1
And now to really muddy the waters !!
One might be expecting the trend to continue when ivsize is 4, and NVprecision is 53bits .... but not so:
C:\_32>perl wle "$m = (2**53)  1; $n = 2; print $m % $n;"
1
C:\_32>perl MPOSIX wle "$m = (2**53)  1; $n = 2; print fmod($m, $n)
+;"
1
This time the "%" operator decides to act in accordance with (my reading of) the perlop documentation.
Cheers,
Rob

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