Matlab中具有双精度的算术精度 | Matlab中具有双精度的算术精度

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问题描述

我有一点麻烦，了解这些双精度如何影响Matlab中算术运算的结果。我以为，既然a& b是双倍的，他们将能够进行这样的精确度的操作。我意识到可能会出现四舍五入的错误，但是由于这些数字在64位数字表示之内，我认为这不是一个问题。

  a = 1.22e-45 
b = 1 
a == 0 
 ans = 0％a不等于零
（a + b）= = 1 
 ans = 1

如何能够承载足够的精度来识别！= 0，但是当被添加到1时，它不会显示任何变化。

解决方案

浮动精确度相对于数字本身的比例。

在给出的具体示例中，1.22e-45可以单独表示，因为指数可以调整为代表10 ^ -45或大约2 ^ -150。

另一方面，1.0以二进制表示，标度为2 ^ 0（即1） p>

要添加这两个值，您需要对齐其小数点（呃...二进制点），这意味着所有精度为1.22e-45当然，IEEE双精度浮点值只有53位尾数（精度），这意味着在1.0，1.22e-45实际上是零。

I am having a bit of trouble understanding how the precision of these doubles affects the outcome of arithmetic operations in Matlab. I thought that since both a & b are doubles they would be able to carry out operations up to that precision. I realize there can be round-off error but since these numbers are well within the 64-bit number representation I didn't think that would be an issue.

a = 1.22e-45
b = 1
a == 0
   ans = 0  %a is not equal to zero
(a + b) == 1
   ans = 1

How come it is able to carry enough precision to recognize a != 0 but when added to 1 it doesn't show any change.

解决方案

"Floating" point means just that--the precision is relative to the scale of the number itself.

In the specific example you gave, 1.22e-45 can be represented alone because the exponent can be adjusted to represent 10^-45, or approximately 2^-150.

On the other hand, 1.0 is represented in binary with scale 2^0 (i.e., 1).

To add these two values, you need to align their decimal points (er...binary points), meaning that all of the precision of 1.22e-45 is shifted 150-odd bits to the right.

Of course, IEEE double precision floating point values only have 53 bits of mantissa (precision), meaning that at the scale of 1.0, 1.22e-45 is effectively zero.

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