英文:
Weird subtraction result in golang
问题
代码中的问题是由于浮点数精度导致的。在Go语言中，浮点数的精度是有限的，因此在进行浮点数计算时可能会出现舍入误差。在这种情况下，idle和busy的值非常大，相加后的结果超出了浮点数的表示范围，导致舍入误差，最终结果为0.000000。
要解决这个问题，可以考虑使用更高精度的数据类型，如big.Float，或者重新设计计算逻辑以避免精度问题。
英文:
code:
func main() {
var idle float64 = 1.8483711496702958e+18
busy := float64(100.0) - idle
sum := idle + busy
fmt.Printf("idle: %f\n", idle)
fmt.Printf("busy: %f\n", busy)
fmt.Printf("sum: %f\n", sum)
}
result:
idle: 1848371149670295808.000000
busy: -1848371149670295808.000000
sum: 0.000000
I would ask why busy + idle = 0.000?
Go Playground: https://go.dev/play/p/WmtK9AuN_yt
答案1
得分: 2
idle相对于busy要大几个数量级，差异相当于10的16次方倍！这比浮点数的精度还要大，浮点数的精度由IEEE 754标准定义。将100加到-idle上不会改变它的值太多。
如果你减小idle，你会得到你期望的结果：
var idle float64 = 1.8483711496702958e+15
通过这个改变，输出将会是（在Go Playground上试一试）：
idle: 1848371149670295.750000
busy: -1848371149670195.750000
sum: 100.000000
浮点数只存储一个数的最重要的位数。如果idle大约是10的18次方，将100加到它上面只会改变那些没有被准确（或根本没有）表示的位数。如果你将idle减小到10的15次方或更低，那么加上100将会改变仍然被表示的位数，你将会看到期望的结果。
英文:
idle compared to busy is several orders of magnitude bigger, the difference is like 10^16 times! That's bigger than the precision of floating point numbers which is defined by IEEE 754 standard. Adding 100 to -idle won't change it "much".
If you decrease idle, you get what you expect:
var idle float64 = 1.8483711496702958e+15
With this change, output will be (try it on the Go Playground):
idle: 1848371149670295.750000
busy: -1848371149670195.750000
sum: 100.000000
Floating point numbers only store the most significant bits of a number. If idle is around 10^18, adding 100 to it would only change bits that are not represented precisely (or at all). If you decrease idle to like 10^15 or lower, then adding 100 will change bits that are still represented, and you'll see the expected result.
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