Golang 实现md5sum 分片计算大文件md5 值
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
package main
import (
"crypto/md5"
"fmt"
"io"
"math"
"os"
)
const fileChunk = 8192 // 8KB
func countFileMd5(filePath string) string {
file, err := os.Open(filePath)
if err != nil {
return err.Error()
}
defer file.Close()
info, _ := file.Stat()
fileSize := info.Size()
blocks := uint64(math.Ceil(float64(fileSize) / float64(fileChunk)))
hash := md5.New()
for i := uint64(0); i < blocks; i++ {
blockSize := int(math.Min(fileChunk, float64(fileSize-int64(i*fileChunk))))
buf := make([]byte, blockSize)
file.Read(buf)
io.WriteString(hash, string(buf))
}
return fmt.Sprintf("%x", hash.Sum(nil))
}
func main() {
if len(os.Args) == 1 {
fmt.Println("run with one or more file path")
fmt.Println("eg: md5sum a.txt b.txt c.iso")
return
}
for i := 1; i < len(os.Args); i++ {
fmt.Println(countFileMd5(os.Args[i]), os.Args[i])
}
}
本文网址: https://golangnote.com/topic/214.html 转摘请注明来源