Question

I m quite new to Go coming from C++ background and have stumbled upon some weird issue. Here s the code:

package main

import (
"fmt"
"unsafe"
)

func main() {
     arr := []string { "one", "two", "three" }
     address := unsafe.Pointer(&arr)
     addPtr := (*[]string)(unsafe.Pointer(*(*uintptr)(address)))
     fmt.Println((*addPtr)[0])
}

此代码失败的原因如下：

运行时错误：growtslice:len超出范围

例如，如果我将演员阵容更改为：

addPtr := (*[0]string)(unsafe.Pointer(*(*uintptr)(address)))

上面的代码工作得很好。

I understand that this is a cast to an array pointer and array must have constant size, but how to cast it to pointer to a slice then?

更令人困惑的是，可以采用切片的地址并将其分配给指针，如下所示：

func main() {
     arr := []string { "one", "two", "three" }
     var arrPtr *[]string = &arr
     fmt.Println((*arrPtr)[0])
}

And this time everything will work despite the fact that the type of pointer is the same type that i was casting unsafe pointer to in the first example. Can someone help to understand what s exactly going on here?

Answer 1

一些背景：切片标头包含指向支持数组、长度和容量的指针。

问题第一部分中的代码将切片标头转换为指向切片标头的指针。go vet命令警告问题中的代码可能被滥用了unsafe.Pointer。

修复方法是删除额外的取消引用操作，使代码从指针到切片标头转换为指针到切片标题。

arr := []string{"one", "two", "three"}
address := unsafe.Pointer(&arr)
addPtr := (*[]string)(unsafe.Pointer((*uintptr)(address)))
fmt.Println((*addPtr)[0]) // prints one

不需要转换为*uintptr。简化为：

arr := []string{"one", "two", "three"}
address := unsafe.Pointer(&arr)
addPtr := (*[]string)(unsafe.Pointer(address))
fmt.Println((*addPtr)[0]) // prints one

不安全的恶作剧是不必要的。简化为：

arr := []string{"one", "two", "three"}
addPtr := &arr
fmt.Println((*addPtr)[0]) // prints one

使用以下代码将切片的后备数组指针转换为数组指针。代码是脆弱的，因为它假定切片标头的第一个字是指向支持数组的指针。

arr := []string{"one", "two", "three"}
address := unsafe.Pointer(&arr)
addPtr := (*[1]string)(unsafe.Pointer(*(*uintptr)(address)))
fmt.Println((*addPtr)[0]) // prints one

不需要前面代码段中的uintptr转换。简化为：

arr := []string{"one", "two", "three"}
address := unsafe.Pointer(&arr)
addPtr := (*[]string)(address)
fmt.Println((*addPtr)[0]) // prints one

我希望这能有所帮助。

友情链接