Go language implementation bitmap sorting (bitmap)

This is a creation in Article, where the information may have evolved or changed.

The go language provides a byte type with a byte corresponding to 8 bits, so the bitmap can be implemented by converting it.

Code:

Package Main//author:xcl//date:2014-1-25import ("FMT") func main () {arrInt32: = [...] Uint32{5, 4, 2, 1, 3, 13}var arrmax uint32 = 20bit: = Newbitmap (Arrmax) for _, V: = range ArrInt32 {bit. Set (v)}fmt. Println ("After sorting:") for I: = UInt32 (0); i < Arrmax; i++ {if k: = bit. Test (i); K = = 1 {fmt. Printf ("%d", i)}}}const (bitsize = 8//One byte 8-bit) type Bitmap struct {BitArray []bytearraysize uint32}func newbitmap (Max UIn T32) *bitmap {var r uint32switch {case Max <= bitsize:r = 1default:r = Max/bitsizeif Max%bitsize! = 0 {r + = 1}}fmt. PRINTLN ("Array size:", R) return &bitmap{bitarray:make ([]byte, R), Arraysize:r}}func (Bitmap *bitmap) Set (i uint32) {idx, P OS: = Bitmap.calc (i) bitmap. BITARRAY[IDX] |= 1 << posfmt. Println ("Set () value=", I, "idx=", idx, "pos=", POS, bytetobinarystring (bitmap. BITARRAY[IDX])}func (bitmap *bitmap) Test (i UInt32) byte {idx, pos: = Bitmap.calc (i) return bitmap. BITARRAY[IDX] >> Pos & 1}func (bitmap *bitmap) Clear (i uint32) {idx, pos: = Bitmap.calc (i) bitMap. BITARRAY[IDX] &^= 1 << pos}func (bitmap *bitmap) calc (i UInt32) (idx, pos uint32) {idx = i >> 3//equals I/8 , that is, the byte position if idx >= bitmap. ArraySize {Panic ("array out of bounds.") Return}pos = i% bitsize//bit position return}//bytetobinarystring function Source://Go language version of the binary string representation of the byte variable//http://www.sharejs.com/codes/ Go/4357func bytetobinarystring (data byte) (str string) {var a bytefor I: = 0; i < 8; i++ {a = Datadata <<= 1data >>= 1switch A {case data:str + = "0" Default:str + = "1"}data <<= 1}return str}

Operation Result:

Array size: 3set ()  value= 5  idx= 0  pos= 5 00100000set ()  value= 4  idx= 0  pos= 4 00110000set ()  Value= 2  idx= 0  pos= 2 00110100set ()  value= 1  idx= 0  pos= 1 00110110set ()  value= 3  idx= 0  pos= 3 00111110set ()  value=  idx= 2  pos= 1 00000010set ()  value= idx=  1  pos= 5 00100000 after sequencing: 1 2 3 4 5 13 17

Mail:xcl_168@aliyun.com

blog:http://blog.csdn.net./xcl168