英文:
Comparing floats by ignoring last bit in golang
问题
以下是要翻译的内容：
一个规范如下所示：
> 如果两个实数在最后一个二进制位上不同，它仍然认为它们相等。
我想在Go语言中为float64数据类型实现这种比较浮点数的方式。不幸的是，位运算符对浮点数没有定义。有没有办法在Go语言中实现这种比较浮点数的方式？
英文:
A specification reads as follows:
> It still considers real numbers equal if they differ in their last
> binary digit.
I would like to implement this way of comparing floats for the float64 data type in Go. Unfortunately, the bitwise operators aren't defined for floating point numbers. Is there a way to achieve this way of comparing floats in the Go language?
答案1
得分: 4
这看起来是使用 math 包中的以下函数的完美案例：
func equal(x, y float64) bool {
return math.Nextafter(x, y) == y
}
Nextafter 函数返回 x 向 y 方向的下一个可表示的 float64 值。
特殊情况包括：
Nextafter(x, x) = x
Nextafter(NaN, y) = NaN
Nextafter(x, NaN) = NaN
链接：https://play.golang.org/p/unRkkoe6wb
英文:
This looks like a perfect use case for the following function from the math package:
func equal(x, y float64) bool {
return math.Nextafter(x, y) == y
}
> Nextafter returns the next representable float64 value after x towards y.
> <br><br>Special cases are:
> <br>Nextafter(x, x) = x
> <br>Nextafter(NaN, y) = NaN
> <br>Nextafter(x, NaN) = NaN
答案2
得分: 2
如果你想知道两个float64值是否相邻（即它们之间没有其他的float64值），可以使用以下代码：
func almostEqual(a, b float64) bool {
ai, bi := int64(math.Float64bits(a)), int64(math.Float64bits(b))
return a == b || -1 <= ai-bi && ai-bi <= 1
}
大多数情况下，这意味着它们在尾数的最低位上有差异。
这段代码在a或b为NaN、零或无穷大时无法正常工作，但你可以根据需要添加特殊情况。
参考链接：https://randomascii.wordpress.com/2012/01/23/stupid-float-tricks-2/
英文:
If you want to know if two float64 values are adjacent (that is, there's no float64 value between them):
func almostEqual(a, b float64) bool {
ai, bi := int64(math.Float64bits(a)), int64(math.Float64bits(b))
return a == b || -1 <= ai-bi && ai-bi <= 1
}
Mostly that's the same as saying they differ in the lowest bit of their mantissa.
This code doesn't work if a or b are NaNs, zeros or infinities, but you could add special cases if you wished.
See https://randomascii.wordpress.com/2012/01/23/stupid-float-tricks-2/
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