如何在Go中将[4]uint8转换为uint32？
如何在Go中将[4]uint8转换为uint32？
提问于 2011-09-12 01:52:32
如何将go的类型从uint8转换为unit32？
只需编写代码：
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测试测试6 && ./test
test.go:10:无法将uInt8 (类型[]uint8)转换为类型uint32
回答 2
Stack Overflow用户
回答已采纳
发布于 2011-09-12 02:26:54
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)
}输出：
little-endian: [0 1 2 3] to 50462976
big-endian: [0 1 2 3] to 66051Go binary package函数以一系列移位的形式实现。
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
}Stack Overflow用户
发布于 2011-09-13 06:37:07
您是否正在尝试使用following
t := []int{1, 2, 3, 4}
s := make([]interface{}, len(t))
for i, v := range t {
s[i] = v
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/7380158
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