英文:
How to convert [4]uint8 into uint32 in Go?
问题
如何将Go语言中的类型从uint8转换为uint32？
只需修改代码如下：
package main
import (
"fmt"
)
func main() {
uInt8 := []uint8{0,1,2,3}
var uInt32 uint32
uInt32 = uint32(uInt8[0])
fmt.Printf("%v to %v\n", uInt8, uInt32)
}
~>6g test.go && 6l -o test test.6 && ./test
test.go:10: 无法将uInt8（类型为[]uint8）转换为类型uint32
英文:
how to convert go's type from uint8 to unit32?<br/>
Just code:
package main
import (
"fmt"
)
func main() {
uInt8 := []uint8{0,1,2,3}
var uInt32 uint32
uInt32 = uint32(uInt8)
fmt.Printf("%v to %v\n", uInt8, uInt32)
}
~>6g test.go && 6l -o test test.6 && ./test <br/>
test.go:10: cannot convert uInt8 (type []uint8) to type uint32
答案1
得分: 16
func main() {
u8 := []uint8{0, 1, 2, 3}
u32LE := binary.LittleEndian.Uint32(u8)
fmt.Println("little-endian:", u8, "to", u32LE)
u32BE := binary.BigEndian.Uint32(u8)
fmt.Println("big-endian: ", u8, "to", u32BE)
}
英文:
package main
import (
"encoding/binary"
"fmt"
)
func main() {
u8 := []uint8{0, 1, 2, 3}
u32LE := binary.LittleEndian.Uint32(u8)
fmt.Println("little-endian:", u8, "to", u32LE)
u32BE := binary.BigEndian.Uint32(u8)
fmt.Println("big-endian: ", u8, "to", u32BE)
}
Output:
little-endian: [0 1 2 3] to 50462976
big-endian: [0 1 2 3] to 66051
The Go binary package functions are implemented as a series of shifts.
func (littleEndian) Uint32(b []byte) uint32 {
return uint32(b[0]) | uint32(b[1])<<8 | uint32(b[2])<<16 | uint32(b[3])<<24
}
func (bigEndian) Uint32(b []byte) uint32 {
return uint32(b[3]) | uint32(b[2])<<8 | uint32(b[1])<<16 | uint32(b[0])<<24
}
答案2
得分: -1
你是否试图实现以下代码？
t := []int{1, 2, 3, 4}
s := make([]interface{}, len(t))
for i, v := range t {
s[i] = v
}
英文:
Are you trying to the following?
t := []int{1, 2, 3, 4}
s := make([]interface{}, len(t))
for i, v := range t {
s[i] = v
}
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