# How does the mod function compensate for floating point round off?

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##### 0 Comments

### Accepted Answer

Joel Handy
on 20 Sep 2022

##### 2 Comments

Bobby Cheng
on 20 Sep 2022

Two things:

1) The compensation only happens when the second input is not of integer values. That is why mod(1+eps,1) returns eps. Otherwise it is what Joel said.

2) I can see there is room for improvement for MATLAB documentation to better clarify this.

### More Answers (3)

Matt J
on 19 Sep 2022

Edited: Matt J
on 19 Sep 2022

I suspect it is just meant to tell you that it doesn't do a naive computation like in modNaive below.

m=1000*pi; n=pi;

mod(m,n)

modNaive(m,n)

It isn't something you should try to rely on, though. There are definitely cases where the result won't be an exact integer, even when it's clearly what would be ideal, e.g.,

mod(117*sqrt(1001)+1,sqrt(1001))-1

function out=modNaive(m,n)

out=m-floor(m/n)*n;

end

##### 6 Comments

Matt J
on 20 Sep 2022

I 'm not sure I can agree, mod must be identical sequence of instructions based on IEEE 754 division and remainder. Any CPU architecture should give identical result.

That may apply to the operations executed within mod itself. However, the preceding computations that generated the inputs to mod may have different floating point errors affecting them. That would be enough to generate different results in the neighborhood of mod's discontinuities.

Bruno Luong
on 20 Sep 2022

Edited: Bruno Luong
on 20 Sep 2022

I try to replicate MATLAB mod ith this function, it seems working well for 2 examples, no warranty beside that.

x=(rand(1,1e6)-0.5)*(10*pi);

all(rmod(x,pi) == mod(x,pi))

x = (-1000:1000)*pi;

all(rmod(x,pi) == mod(x,pi))

function r = rmod(x, a)

k = round(x/a);

r = x - a*k;

r(abs(r) < eps(x)) = 0; % EDIT test probably non effective

r(r < 0) = r(r < 0) + a;

end

##### 0 Comments

Bruno Luong
on 20 Sep 2022

Second version, correction under stricter condition than in the first version

function r = rmod2(x, a)

k = round(x/a);

r = x - a*k;

r(abs(r) < eps(a)) = 0;

r(r < 0) = r(r < 0) + a;

end

##### 1 Comment

Bruno Luong
on 20 Sep 2022

I think my test with espsilon is actually useless, this seem to match mod, unless someone can come with a counter example

function r = rmod(x, a)

k = round(x/a);

r = x - a*k;

r(r < 0) = r(r < 0) + a;

end

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