解决Golang小数float64在实际工程中加减乘除的精度问题

var (
a, b float64
c int64
)
a = 2.55
b = 0.0

c = int64((a + b) * 100.0)
fmt.Printf("第1次 c=%d\n", c) //第1次 c=254
c = int64(a * 100.0)
fmt.Printf("第2次 c=%d\n", c) //第2次 c=254
fmt.Println(a * 100) //254.99999999999997
//发现将小数的元乘以100后强制转换为整数分，少了1分
//解决办发1：
tmpStr1 := fmt.Sprintf("%.2f", a)
tmpStr2 := fmt.Sprintf("%.2f", b)
tmpnum1, _ := strconv.ParseInt(strings.Replace(tmpStr1, ".", "", 1), 10, 64)
tmpnum2, _ := strconv.ParseInt(strings.Replace(tmpStr2, ".", "", 1), 10, 64)
c = tmpnum1 + tmpnum2
fmt.Printf("第3次 c=%d\n", c) //第3次 c=255

num3 := math.Ceil(a * 100.0)
fmt.Printf(“num3=%d\n”, int64(num3)) //num3=255
//方法3：
strFloat := strconv.FormatFloat(a*100, ‘f', 0, 64)
nInt64, err := strconv.ParseInt(strFloat, 10, 64)

//将float64转成精确的int64
func Wrap(num float64, retain int) int64 {
 return int64(num * math.Pow10(retain))
}
//将int64恢复成正常的float64
func Unwrap(num int64, retain int) float64 {
 return float64(num) / math.Pow10(retain)
}
//精准float64
func WrapToFloat64(num float64, retain int) float64 {
 return num * math.Pow10(retain)
}
//精准int64
func UnwrapToInt64(num int64, retain int) int64 {
 return int64(Unwrap(num, retain))
}