英文:
How to convert int[] to uint8[]
问题
所以,我需要你的帮助。我在那个主题上找不到任何东西。Golang是一种新近开发的语言,所以对于像我这样的新手来说,很难快速找到答案。
英文:
SO, I need your help. I couldn’t find anything on that topic. Golang is a freshly baked language so it’s quite hard to find answers quick for a newcomers like me.
答案1
得分: 5
预定义的Go int
类型的大小是与实现相关的,可以是32位或64位(数字类型)。
这里有一个将大端序的int
转换为byte
(uint8
)的示例。
package main
import (
"encoding/binary"
"fmt"
"reflect"
)
func IntsToBytesBE(i []int) []byte {
intSize := int(reflect.TypeOf(i).Elem().Size())
b := make([]byte, intSize*len(i))
for n, s := range i {
switch intSize {
case 64 / 8:
binary.BigEndian.PutUint64(b[intSize*n:], uint64(s))
case 32 / 8:
binary.BigEndian.PutUint32(b[intSize*n:], uint32(s))
default:
panic("unreachable")
}
}
return b
}
func main() {
i := []int{0, 1, 2, 3}
fmt.Println("int size:", int(reflect.TypeOf(i[0]).Size()), "bytes")
fmt.Println("ints:", i)
fmt.Println("bytes:", IntsToBytesBE(i))
}
输出:
int size: 4 bytes
ints: [0 1 2 3]
bytes: [0 0 0 0 0 0 0 1 0 0 0 2 0 0 0 3]
或者
int size: 8 bytes
ints: [0 1 2 3]
bytes: [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 3]
英文:
The predeclared Go int
type size is implementation-specific, either 32 or 64 bits (Numeric types).
Here's an example of converting big-endian int
s to byte
s (uint8
s).
package main
import (
"encoding/binary"
"fmt"
"reflect"
)
func IntsToBytesBE(i []int) []byte {
intSize := int(reflect.TypeOf(i).Elem().Size())
b := make([]byte, intSize*len(i))
for n, s := range i {
switch intSize {
case 64 / 8:
binary.BigEndian.PutUint64(b[intSize*n:], uint64(s))
case 32 / 8:
binary.BigEndian.PutUint32(b[intSize*n:], uint32(s))
default:
panic("unreachable")
}
}
return b
}
func main() {
i := []int{0, 1, 2, 3}
fmt.Println("int size:", int(reflect.TypeOf(i[0]).Size()), "bytes")
fmt.Println("ints:", i)
fmt.Println("bytes:", IntsToBytesBE(i))
}
Output:
int size: 4 bytes
ints: [0 1 2 3]
bytes: [0 0 0 0 0 0 0 1 0 0 0 2 0 0 0 3]
or
int size: 8 bytes
ints: [0 1 2 3]
bytes: [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 3]
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