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About floor function problem.

Asked by C Zeng on 23 Oct 2012

floor(1.999999999999)=1 floor(1.99999999999999999999999999)=2, why is that?

Floor should return the lower integer right? Thanks.

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C Zeng

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2 Answers

Answer by Matt J on 23 Oct 2012
Edited by Matt J on 23 Oct 2012
Accepted answer

If that confuses you, this probably will too:

>> isequal(1.99999999999999999999999999, 2)
ans =
       1

Anyway, it has nothing to do with the FLOOR command. It's because your big long decimal can't be distinguished from 2 in floating point.

6 Comments

C Zeng on 23 Oct 2012

Thanks a lot Matt!

Would you please answer my previous question?

Matt J on 23 Oct 2012

What was your "previous question"?

Matt J on 23 Oct 2012

This one contains an overloaded floor function, if that's what you mean

http://www.mathworks.com/matlabcentral/fileexchange/6446-multiple-precision-toolbox-for-matlab

Matt J
Answer by Azzi Abdelmalek on 23 Oct 2012

Just try without floor

a=1.99999999999999999999999999

8 Comments

Azzi Abdelmalek on 23 Oct 2012

I am not sur what you mean by converting to 2 digits, I think, with 2 digits, you will have four possible digital numbers, And you need a min and max value to be able to do this conversion. Can you explain, or post another question?

Walter Roberson on 23 Oct 2012

If you are starting with an integer, then dividing by a power of 2 can never result in this kind of round-off. Powers of 2 are represented exactly in binary floating point numbers, and dividing by a power of two effectively only changes the internal binary exponent without changing the mantissa. If you are running into this kind of round-off then either you are not starting with an integer or you are not dividing by a power of 2.

C Zeng on 26 Oct 2012

Thanks, Walter, though I do not understand your point. I am transferring an integer like N to 2-digits. I want to divide it by 2 to determine if the entry is 0 or 1. Floor function does not make sufficient proximity to this problem.

Azzi Abdelmalek

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