golang十进制到十六进制转换错误

Question

Problem with precision

The setup is a physical i-button (1-wire specification) with printed hex value by manufacturer (trusted value). An IoT device encodes the binary as decminal - Golang returns the received value to Hex.

received decimal value is:

var V = 10736581604118680000
fmt.Sprintf("%016X", m.V)[2:14]) // adds uppercase and truncation of output

provides 000015877CD1

The expected trusted output is 000015877CD0

the engraved Hex on the key is 95 000015877CD0 01

http://www.rapidtables.com/convert/number/decimal-to-hex.htm (trusted?) indicates that the golang function used has lost precision. binary values that encode to 19 decimal digits can be converted to Hex by Golang without loss of precision (using function above)

Answer 1

For example,

package main

import (
    "fmt"
    "math/big"
    "strconv"
)

func hexdec(s string) uint64 {
    d := uint64(0)
    for i := 0; i < len(s); i++ {
        x := uint64(s[i])
        if x >= 'a' {
            x -= 'a' - 'A'
        }
        d1 := x - '0'
        if d1 > 9 {
            d1 = 10 + d1 - ('A' - '0')
        }
        if 0 > d1 || d1 > 15 {
            panic("hexdec")
        }
        d = (16 * d) + d1
    }
    return d
}

func main() {
    x := "95000015877CD001"
    fmt.Println(x)
    n, err := strconv.ParseUint(x, 16, 64)
    fmt.Println(n, err)
    s := fmt.Sprintf("%016X", n)[2:14]
    fmt.Println(s)
    z, t := big.NewInt(0).SetString(x, 16)
    fmt.Println(z, t)
    s = fmt.Sprintf("%016X", z)[2:14]
    fmt.Println(s)
    fmt.Println(hexdec(x))
}

Output:

95000015877CD001
10736581604118679553 <nil>
000015877CD0
10736581604118679553 true
000015877CD0
10736581604118679553

Note that you are near the limits of 64-bit integers:

uint64  the set of all unsigned 64-bit integers (0 to 18446744073709551615)

Where does

Var V = 10736581604118680000

come from?

---

The number 10736581604118680000 is a floating-point approximation (1.073658160411868e+19) of the integer 10736581604118679553. The manufacturer probably doesn't understand floating-point: What Every Computer Scientist Should Know About Floating-Point Arithmetic. Given the integer 10736581604118680000, Go calculates the correct result. Go is mathematically correct.

So let's try telling Go that 10736581604118680000 is not an exact integer, that it's an approximate floating-point number.

For example,

package main

import (
    "fmt"
    "strconv"
)

func main() {
    d := "10736581604118680000"
    f, err := strconv.ParseFloat(d, 64)
    fmt.Println(f, err)
    z := uint64(f)
    fmt.Println(z)
    s := fmt.Sprintf("%016X", z)[2:14]
    fmt.Println(s)
}

Output:

1.073658160411868e+19 <nil>
10736581604118679552
000015877CD0

While this trick works in this case, it's not guaranteed to work in all cases. The real solution is for the manufacturer to employ some competent mathematicians. I wonder what other bugs there are in their software and hardware.