Golang binary Packet--byte How does an array go int?

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In the C language test, I prefer to test this thing, how to convert a char array into an int type. I've seen it, but I've forgotten it. Later saw this kind binary.BigEndian.Uint32(a) of thing, straight blind. Later went to read the document, looked at half a day also did not understand.

Here directly to say, source code. The following is the code that is the uint8 byte array, size 4, converted to Int32.

package mainimport "fmt"import "encoding/binary"func main() {var a []byte = []byte{0, 1, 2, 3}fmt.Println(a)fmt.Println(binary.BigEndian.Uint32(a))fmt.Println(binary.LittleEndian.Uint32(a))}

Execution Result:

[0 1 2 3]6605150462976

There are two different ways to convert, that is, the big and small ends. The big-endian is the low address in memory corresponding to the high number of integers. According to the above example, it is spelled in 0123 order int32 , the highest 8 bits of the integer is 0, then 1, and so on, so 66051. The small end is the reverse, the highest 8 bits is 3, that is 00000101 , this finally gets 50462976.

In the Golang source directory encoding/binary.go , there is the implementation of the function. However, there is no check on the length of the byte array, if the incoming array length is less than 4, the error will naturally be: panic: runtime error: index out of range .

func (bigEndian) Uint32(b []byte) uint32 {return uint32(b[3]) | uint32(b[2])<<8 | uint32(b[1])<<16 | uint32(b[0])<<24}

The above BigEndian and LittleEndian all are implementation classes, are implemented classes ByteOrder .

type ByteOrder interface {    Uint16([]byte) uint16    Uint32([]byte) uint32    Uint64([]byte) uint64    PutUint16([]byte, uint16)    PutUint32([]byte, uint32)    PutUint64([]byte, uint64)    String() string}

###### References + "1" byte order (Endian), Big End (Big-endian), Small (Little-endian)-led by wife Full Street + "2" Package binary-the Go Programming Language

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