improve remainder and modulo exact---inexact combinations

[mflatt]

Intended to fix #1008:

• When the second argument is 1 or -1 and the first argument is inexact, return inexact 0.0.

• When the first argument is exact and the second argument is inexact, compute as exact and then convert to inexact.

A calculation like (remainder #e1e1000 3.0) is about 10 times slower. Interestingly, (remainder #e1e100 3.0) is about 10 times faster, and (remainder #e1e10 3.0) is about twice as fast.

[burgerrg]

For the case of remainder with exact 1 or -1, why would the answer be inexact? The remainder of any real number and exact 1 is exactly 0.

The update to the exact/inexact case produces the closest correctly rounded double, which makes a lot of sense to me.

[mflatt]

For the case of remainder with exact 1 or -1, why would the answer be inexact? The remainder of any real number and exact 1 is exactly 0.

I don't think that's the way remainder and modulo are usually generalized to rationals. For example, 1.2 % 1 in Python is close to 0.2. That works out differently than, say, multiplying by exact 0, where the result does always end up being 0.

We're not generalizing to rationals, but an exact 0 result still seems right to me on the grounds that something like 7.0 is not exactly an integer.

[burgerrg]

Thanks, I now understand the rationale for making the remainder of an inexact integer with 1 or -1 to be inexact 0.0. That makes sense.