2014-06-05 23:48
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Golang AES ECB加密

Trying to emulate an algorithm in Go that is basically AES ECB Mode encryption.

Here's what I have so far

func Decrypt(data []byte) []byte {
    cipher, err := aes.NewCipher([]byte(KEY))
    if err == nil {
        cipher.Decrypt(data, PKCS5Pad(data))
        return data
    return nil

I also have a PKCS5Padding algorithm, which is tested and working, which pads the data first. I cant find any information on how to switch the encryption mode in the Go AES package (it's definitely not in the docs).

I have this code in another language, which is how I know this algorithm isn't working quite correctly.

EDIT: Here is the method as I have interpreted from on the issue page

func AESECB(ciphertext []byte) []byte {
    cipher, _ := aes.NewCipher([]byte(KEY))
    fmt.Println("AESing the data")
    bs := 16
    if len(ciphertext)%bs != 0     {
        panic("Need a multiple of the blocksize")

    plaintext := make([]byte, len(ciphertext))
    for len(plaintext) > 0 {
        cipher.Decrypt(plaintext, ciphertext)
        plaintext = plaintext[bs:]
        ciphertext = ciphertext[bs:]
    return plaintext

This is actually not returning any data, maybe I screwed something up when changing it from encripting to decripting

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4条回答 默认 最新

  • doumanni3501 2017-01-14 16:59

    ESB is a very straightforward mode of operation. The data to be encrypted is divided into byte blocks, all having the same size. For each block, a cipher is applied, in this case AES, generating the encrypted block.

    The code snippet below decrypts AES-128 data in ECB (note that the block size is 16 bytes):

    package main
    import (
    func DecryptAes128Ecb(data, key []byte) []byte {
        cipher, _ := aes.NewCipher([]byte(key))
        decrypted := make([]byte, len(data))
        size := 16
        for bs, be := 0, size; bs < len(data); bs, be = bs+size, be+size {
            cipher.Decrypt(decrypted[bs:be], data[bs:be])
        return decrypted

    As mentioned by @OneOfOne, ECB is insecure and very easy to detect, as repeated blocks will always encrypt to the same encrypted blocks. This Crypto SE answer gives a very good explanation why.

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