如何在Golang中将float64转换为c99十六进制字符串？

Question

How to convert float64 to hex string(that follows c99 standard) in Golang?

-561.2863 to -0x1.18a4a57a786c2p9

Answer 1

For example,

package main

import (
    "fmt"
    "math/big"
)

func main() {
    f := new(big.Float).SetFloat64(-561.2863)
    fmt.Println(f)
    t := f.Text('p', 0)
    fmt.Println(t)
    g, ok := new(big.Float).SetString(t)
    if !ok {
        fmt.Println("error")
    }
    fmt.Println(g)
    c, ok := new(big.Float).SetString("-0x1.18a4a57a786c2p9")
    if !ok {
        fmt.Println("error")
    }
    fmt.Println(c)
}

Playground: https://play.golang.org/p/9BygKkJcNbm

Output:

-561.2863
-0x.8c5252bd3c361p+10
-561.2862999999999829
-561.2862999999999829

---

Or, simply,

package main

import (
    "fmt"
    "math/big"
)

func main() {
    f64 := -561.2863
    fmt.Println(f64)
    s := new(big.Float).SetFloat64(f64).Text('p', 0)
    fmt.Println(s)
}

Playground: https://play.golang.org/p/MR2AxtlBcqv

Output:

-561.2863
-0x.8c5252bd3c361p+10

---

In C,

#include <stdio.h>

int main() {
    double f = -561.2863;
    printf("%f %a
", f, f);
}

Output:

-561.286300 -0x1.18a4a57a786c2p+9