http://www.perlmonks.org?node_id=11133829

syphilis has asked for the wisdom of the Perl Monks concerning the following question:

Hi,

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 NV-precision is greater than IV-precision.
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, NV-precision 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 floating-point 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, NV-precision 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, NV-precision 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 NV-precision is 53-bits .... 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