Question

如果我将int数存储到一个结构中，然后将它们应用于它们，那恰好发生在我身上。精度会丢失。

func main() {
   var x = 94911151
   var y = 94911150
   // If we use the value to calculate division directly, it would be fine
   var result1 = float64(94911151)/94911150
   var result2 = float64(x)/float64(y)
   fmt.Println(result1, result2)
   // If we pass the values directly as parameter into a function and then apply division, it would be fine
   getParas(x,y)
   // If we pass the values into a stuct, and then retrieve the value from struct, then apply division, the precision would be lost.
   getLinearParas(Point{x,y},Point{0,0})
}

func getParas(a int, b int){
    diffX := a -0
    diffY := b-0
    c:= float64(diffX) / float64(diffY)
    fmt.Println(c)
}

type Point struct{
    X int
    Y int
}

func getLinearParas(point1 Point, point2 Point)  {
    diffX := point1.X - point2.X
    diffY := point1.Y - point2.Y
    a := float64(diffX) / float64(diffY)
    fmt.Printf("diffY: %d; diffX:%d ; a:%f \n", diffY, diffX, a)
}

与代码一样，如果我将int值放入结构中，稍后对它们应用除法。精度会以某种方式丢失。运行上面代码的结果是

 1.00000001053617 1.00000001053617
 1.00000001053617
 diffY: 94911150; diffX:94911151 ; a:1.000000

或者你可以在游乐场自己尝试一下 https://play.golang.org/p/IDS18rfv9e6

有人可以解释为什么会这样吗？以及如何避免这种损失？非常感谢你。

Answer 1

将格式字符串中的%f更改为%v或%g或%.14f。 fmt.Println打印相当于%v，%v的内容，因为float64被视为%g。默认情况下，%f打印带有6位有效数字的值，%g使用“唯一标识该值所需的最小位数”。

打印时精度损失浮动为字符串

1 个答案: