calculation error for floor function
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Hello all, when I was trying for >>floor(2.3 * 50) , I was getting 114 where I suppose to get 115. Is this due to precision issue? and How to fix it?
1 个评论
Shlomo Geva
2017-10-29
Well, check this code: floor(p/65536) floor() does not handle correctly uint32 values of p > 4294934528 which is well within uint32 range. It is not even returning a value in the expected range 0 to 65535
>> p=uint32([4294934527, 4294934528, 2^32-1]);floor(p/65536)
ans =
1×3 uint32 row vector
65535 65536 65536
回答（2 个）
Sebastian
2017-2-6
If you have the Symbolic Math Toolbox, try this:
x = vpa(2.3 * 50)
x =
115.0
x = floor(x)
x =
115
Jan
2017-2-6
0 个投票
Yes, this is an effect of the limited precision of the IEEE754 floating point standard. See http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F . You find a lot of corresponding question, when you search e.g. for "faq 6.1".
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2017-2-6
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