如何将[] byte转换为[8] uint8
I need to populate a struct that has a member of type [8]uint8
. This needs be populated with a byte array of type []byte
initialized to length 8. The simplistic approach does not work:
Data: [8]uint8(RequestFrame(0x180, r)),
gives
cannot convert .. (type []byte) to type [8]uint8
Since both arrays are structurally identical it would be nice if this could be done with casting/assignment rather than copying?
我需要填充一个成员类型为 给予 p>
由于两个数组在结构上都是相同的,如果可以通过强制转换/赋值而不是复制来完成,这会很好吗? / p>
div> [8] uint8 code>的结构。 这需要使用初始化为长度为8的
[] byte code>类型的字节数组填充。简单的方法不起作用: p>
数据:[8 ] uint8(RequestFrame(0x180,r)),
code> pre>
无法转换..(类型[ ] byte)键入[8] uint8
code> pre>
Background
The problem with your "simplistic approach" is that a slice
(of any type) is a struct
-typed consisting of a pointer
and two integers; the pointer contains the address of the
underlying (backing) data array, and the integers contain
what len()
and cap()
builtins return for that slice.
In other words, a slice is sort of a view into an array.
Then, in Go, there is no concept of a type cast; there are only type conversions, and these conversions may only happen between types with the same underlying representation¹.
Since a slice and an array may not have the same underlying representation (array is literally a contiguous block of memory of the size just enough to contain all the array's elements), your alleged type conversion may not be legal.
Possible solutions
There are two possible solutions.
The simplest is to just copy the data from the slice's backing array into a newly-allocated array:
var (
src = []byte{1, 2, 3, 4, 5, 6, 7, 8}
dst [8]uint8
)
copy(dst[:], src[:8])
Note that there exists an inherent disparity between slice an array types: an array type encodes both the type of its elements and its length (that is, the length is a part of the type), while a slice type only encodes the type of its elements (and may be of any length at runtime).
This means that you might need to have a check before such
copying that makes sure the source slice has exactly 8
elements, that is, len(src) == len(dst)
.
This invariant may be enforced by some other code, but I think
I'd warn you up front about this: if src
has less than 8
elements, the src[:8]
expression will panic at runtime,
and if it contains more, then there's the question of whether
copying just the first 8 of them is exactly what's needed.
The second approach (admittedly messier) is to just re-use the underlying array of the slice:
import "unsafe"
var (
src = []byte{1, 2, 3, 4, 5, 6, 7, 8}
dstPtr *[8]uint8
)
if len(src) != len(*dstPtr) {
panic("boom")
}
dstPtr = (*[8]uint8)(unsafe.Pointer(&src[0]))
Here, we've just taken the address of the first element
contained in the slice's underlying array and peformed
a "dirty" two-phase type-conversion to make the obtained
pointer to be of type *[8]uint8
—that is, "an address of
an array of 8 uint8
s".
Note two caveats:
The resulting pointer now points to the same memory block the original slice does. It means it's now possible to mutate that memory both through the slice and the pointer we obtained.
-
As soon as you'll decide to assign the array's data to a variable of type
[8]uint8
(and passing it as an argument to a function's parameter of that type), you will dereference that pointer (like with*dstPtr
), and at that moment the array's data will be copied.I'm specifically mentioning this as often people resort to hacks like this one to pull the backing array out of a slice precisely in an attempt to not copy the memory.
TL;DR
Copy the data (after supposedly verifying the
len(src) == len(dst)
invariant holds).
Copying 8 bytes is fast (on a typical 64-bit CPU this will be
a single MOV
instruction, or two at most), and the code will
be straightforward.
Only resort to hacks from the second solution when you really need to optimize on some critical hot path. In that case, comment the solution extensively and watch for not accidentally dereferencing your pointer.
¹ There is a single but notable exception to this rule:
a []byte
is type-convertible to string
, and vice-versa.
You can copy the contents of your byte
slice into your uint8
array very simply by using copy
, like this:
package main
import (
"fmt"
)
func main() {
slice := []byte{1, 2, 3, 4, 5, 6, 7, 8}
array := [8]uint8{}
copy(array[:], slice)
fmt.Println(array)
}
Outputs
[1 2 3 4 5 6 7 8]
But may I ask why you are using an array? It's usually better to just use slices, unless you have a really good reason.