About floor function problem.

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

Floor should return the lower integer right? Thanks.

Answer 1

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.

Oh, yes, Matt. Then is there a way to adjust the precision in Matlab? Thanks.

There are FEX submissions which support alternative precisions. Here is one example

http://www.mathworks.com/matlabcentral/fileexchange/36534

I tend to agree with what the author says there, however: If you can't do it with ordinary double precision, you're probably doing something wrong.

http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

To get higher precision, use the Symbolic Toolbox, or the Fixed Point Toolbox, or one of the FEX contributions for higher precision numbers.

MATLAB itself is limited by its use of (hardware) IEEE 754 Double Precision numbers.

Thanks a lot Matt!

Would you please answer my previous question?

What was your "previous question"?

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

Answer 2

Just try without floor

a=1.99999999999999999999999999

I have to floor it, Azzi. I want an integer.

I mean you don't need to floor it, you will find 2

I still want to convert 1.9999999999999 to 1. How can I do it?

Zeng. from where did you get 1.999999999999999? knowing that matlab don't allow it

Azzi, I see. I want to convert a number y to 2-digit vector. I use y/2^(i-1) iteratively to convert y to the 2-digit expression. To make sure it display correct 0 or 1, I use floor function. However if the division is very close to 1, floor() will show 1 not 0.

I think I can modify it by comparing who is greater, y and 2^(i-1).

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?

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.

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.