Question

页: 1 缩略语我无法在网上找到解决这一问题的办法。我看到了将第一组星改为“条码”的建议,然后改为“条码>float64,但这似乎行不通,失去价值,我最后说是零。

Example:

metric.Value, _ = strconv.ParseFloat(string(column.Value), 64)

并且不工作......

Answer 1

For example,

package main

import (
    "encoding/binary"
    "fmt"
    "math"
)

func Float64frombytes(bytes []byte) float64 {
    bits := binary.LittleEndian.Uint64(bytes)
    float := math.Float64frombits(bits)
    return float
}

func Float64bytes(float float64) []byte {
    bits := math.Float64bits(float)
    bytes := make([]byte, 8)
    binary.LittleEndian.PutUint64(bytes, bits)
    return bytes
}

func main() {
    bytes := Float64bytes(math.Pi)
    fmt.Println(bytes)
    float := Float64frombytes(bytes)
    fmt.Println(float)
}

Output:

[24 45 68 84 251 33 9 64]
3.141592653589793

Answer 2

I think this example from Go documentation is what you are looking for: http://golang.org/pkg/encoding/binary/#example_Read

var pi float64
b := []byte{0x18, 0x2d, 0x44, 0x54, 0xfb, 0x21, 0x09, 0x40}
buf := bytes.NewReader(b)
err := binary.Read(buf, binary.LittleEndian, &pi)
if err != nil {
  fmt.Println("binary.Read failed:", err)
}
fmt.Print(pi)

印刷 3.141592653589793

Answer 3

如评注所示,所有数据都取决于您在<代码>上拥有何种数据。

如果是代表小端人订单的754个浮点值的中间商,则使用Kluyg 或石块。因此,答案(在不使用反省的情况下提高业绩)。

如果是拉丁-1/UTF-8编码的准文本代表,那么你就应当能够做你刚才做的事情:

package main

import (
    "fmt"
    "strconv"
)

func main() {
    var f float64
    text := []uint8("1.23") // A decimal value represented as Latin-1 text

    f, err := strconv.ParseFloat(string(text), 64)
    if err != nil {
        panic(err)
    }

    fmt.Println(f)
}

结果:

1.23

游乐场:

Answer 4

I hope this hack helps. The purpose of it is to convert the long stream of binary numbers to float.

For example: 0110111100010010100000111100000011001010001000010000100111000000 -> -3.1415

func binFloat(bin string) float64 {

    var s1 []byte
    var result float64

    if len(bin) % 8 == 0 {

            for i := 0; i < len(bin) / 8; i++ {

                    //Chop the strings into a segment with a length of 8.
                    //Convert the string to Integer and to byte

                    num, _ := strconv.ParseInt(bin[8*i: 8*(i + 1)], 2, 64) 
                    //Store the byte into a slice s1
                    s1 = append(s1, byte(num)) 
            }

    }

    //convert the byte slice to a float64. 
    //The algorithm below are copied from golang binary examples. 

    buf := bytes.NewReader(s1)

    //You can also change binary.LittleEndian to binary.BigEndian
    //For the details of Endianness, please google Endianness

    err := binary.Read(buf, binary.LittleEndian, &result) 

    if err != nil {

            panic(err)
            fmt.Println("Length of the binary is not in the length of 8")
    }

    return result
}

Answer 5

至今已经接受的答案只是为了更长的阵列而努力,有8个 by子为短阵。

这还包括少量人员:

bits := big.NewInt(0).SetBytes(raw).Uint64()
float := math.Float64frombits(bits)

友情链接