英文:
All entries in slice end up identical after copying from another slice
问题
我有一个 uuid.UUID
元素的切片。根据源代码,uuid.UUID
只是一个字节数组的类型别名:type UUID [16]byte
。
我试图使用以下代码将这个 uuid.UUID
切片转换为 []byte
切片:
package main
import (
"fmt"
"github.com/google/uuid"
)
func main() {
ids := []uuid.UUID {
uuid.New(),
uuid.New(),
uuid.New(),
}
fmt.Printf("ids: %v\n", ids)
idBytes := make([][]byte, len(ids))
for i, id := range ids {
idBytes[i] = id[:]
}
fmt.Printf("idBytes: %x\n", idBytes)
}
由于某种原因,这会产生以下输出:
ids: [66b4bb26-3c1f-4dd7-a608-aa8f799e4bfe 014e0537-c326-4021-be38-165f49595eed 5c71efff-ddb5-4f6e-8f85-c1dab013b5d1]
idBytes: [5c71efffddb54f6e8f85c1dab013b5d1 5c71efffddb54f6e8f85c1dab013b5d1 5c71efffddb54f6e8f85c1dab013b5d1]
显然有 3 个不同的 UUID,但输出的切片只包含最后一个 UUID,重复了 3 次。这是怎么回事?
我尝试过的方法:
- 使用非切片类型而不是切片类型(即,使用 []uint64 作为输入和输出,而不是 []uuid.UUID 和 [][]byte)。这显示了预期的行为(我看到了 3 个不同的值,而不是最后一个值重复 3 次),但它并没有真正解决我的问题。
- 只设置最后一个元素(将
idBytes[i] = id[:]
放在一个if i == len(ids) - 1
的块中)。这只设置了最后一个元素(前两个元素仍然为 nil)。
英文:
I have a slice of uuid.UUID
elements. From the source, a uuid.UUID
is just a type alias for a byte array: type UUID [16]byte
.
I'm trying to convert this slice of uuid.Uuid
s to a slice of []byte
using the following code:
package main
import (
"fmt"
"github.com/google/uuid"
)
func main() {
ids := []uuid.UUID {
uuid.New(),
uuid.New(),
uuid.New(),
}
fmt.Printf("ids: %v\n", ids)
idBytes := make([][]byte, len(ids))
for i, id := range ids {
idBytes[i] = id[:]
}
fmt.Printf("idBytes: %x\n", idBytes)
}
For some reason, this produces the following output:
ids: [66b4bb26-3c1f-4dd7-a608-aa8f799e4bfe 014e0537-c326-4021-be38-165f49595eed 5c71efff-ddb5-4f6e-8f85-c1dab013b5d1]
idBytes: [5c71efffddb54f6e8f85c1dab013b5d1 5c71efffddb54f6e8f85c1dab013b5d1 5c71efffddb54f6e8f85c1dab013b5d1]
There are clearly 3 distinct UUIDs, but the output slice contains only the last one, repeated 3 times. What's going on here?
Things I've tried:
- Use a non-slice type instead of a slice type (that is, use []uint64 as input and output instead of []uuid.UUID and [][]byte). This shows the expected behavior (I see 3 distinct values instead of the last value repeated 3 times), but it doesn't actually solve my problem.
- Set only the last element (put the
idBytes[i] = id[:]
inside anif i == len(ids) - 1
block). This sets only the last element (the first two elements remain nil).
答案1
得分: 5
考虑以下两个事实:
id
的类型是 UUID,它是[16]byte
类型。- 循环变量在每次迭代时都会被覆盖。
因此,在 for 循环的所有迭代中,只有一个 id
实例被分配和共享。由于 id
是一个数组(而不是切片),每个 UUID 的内容都会被复制到 id
中。所有的切片 id[:]
都指向相同的共享底层字节数组,当循环结束时,该数组中包含了 id
的最新内容。
修复方法如下:
idBytes[i] = ids[i][:]
这将为每个 UUID 创建单独的切片。
英文:
Two facts to consider:
id
is of type UUID, which is[16]byte
.- Loop variables a overwritten at each iteration
Thus, there is only one instance of id
is allocated and shared between all iterations of the for loop. Since id
is an array (not a slice), contents of each UUID are copied to id
. All the slices id[:]
, thus, point to the same shared underlying byte array, which, when the loop ends, contains the latest contents of id
.
To fix:
idBytes[i] = ids[i][:]
which will create separate slices for each UUID.
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