GoLang for float循环会产生错误

Can someone explain the following. I have a function in go that accepts a couple of float64 and then uses this value to calculate a lot of other values. The function looks like

func (g *Geometry) CalcStresses(x, zmax, zmin float64)(Vertical)

the result is put into a struct like

type Vertical struct {
    X float64
    Stresses []Stress
}

Now the funny thing is this. If I call the function like this;

for i:=14.0; i<15.0; i+=0.1{
    result := geo.CalcStresses(i, 10, -10)
}

then I get a lot of results where the Stress array is empty, antoher interesting detail is that x sometimes shows like a number with a LOT of decimals (like 14.3999999999999999998)

However, if I call the function like this;

for i:=0; i<10; i++{
    x := 14.0 + float64(i) * 0.1
    result := geo.CalcStresses(x,10,-10)
}

then everything is fine.

Does anyone know why this happens?

Thanks in advance, Rob

Not all real numbers can be represented precisely in binary floating point format, therefore looping over floating point number is asking for trouble.

From Wikipedia on Floating point

The fact that floating-point numbers cannot precisely represent all real numbers, and that floating-point operations cannot precisely represent true arithmetic operations, leads to many surprising situations. This is related to the finite precision with which computers generally represent numbers.

For example, the non-representability of 0.1 and 0.01 (in binary) means that the result of attempting to square 0.1 is neither 0.01 nor the representable number closest to it.

This code

for i := 14.0; i < 15.0; i += 0.1 {
    fmt.Println(i)
}

produces this

14
14.1
14.2
14.299999999999999
14.399999999999999
14.499999999999998
14.599999999999998
14.699999999999998
14.799999999999997
14.899999999999997
14.999999999999996

You may use math.big.Rat type to represent rational numbers accurately.

Example

x := big.NewRat(14, 1)
y := big.NewRat(15, 1)
z := big.NewRat(1, 10)

for i := x; i.Cmp(y) < 0; i = i.Add(i, z) {
    v, _ := i.Float64()
    fmt.Println(v)
}