如何在Golang中将字符串转换为* uint64？

假设有一个字符串保存类型变量的地址，我们可以将此地址解析成这样吗？uint64 *uint64

例如：

i := uint64(23473824)
ip := &i
str := fmt.Sprintf("%v", ip)

u, _ := strconv.ParseUint(str, 0, 64)

u是uint64，该如何从这个值中取出指针?

背景链接： https://play.golang.org/p/1KXFQcozRk

You can do it with

 ip = (*uint64)(unsafe.Pointer(uintptr(u)))

playground link

Albeit I don't know what guarantees Go gives you about the validity of such a pointer, nor can I think of any use case where this code should be used..

Based on nos answer.

Although it is technically possible there are reasons not to trust the code you wrote. Garbage collection will use the memory you point to (with string).

Take a look at result of the following code.

package main

import(
    "fmt"
    "strconv"
    "reflect"
    "unsafe"
)

func produce() string {
    i := uint64(23473824)
    ip := &i
    str := fmt.Sprintf("%v", ip)
    fmt.Println(i, ip, str)
    return str
}

func main() {
    str := produce()

    for i := 0; i < 10; i++ {
         x := make([]int, 1024*1024)
         x[0] = i
    }        

    u, _ := strconv.ParseUint(str, 0, 64) 

    ip := (*uint64)(unsafe.Pointer(uintptr(u)))
    fmt.Println(ip,*ip, reflect.TypeOf(u)) // u is uint64, how to get pointer out of this value?
}

https://play.golang.org/p/85XOhsMTf3