Question

golang rc4 结果不正确.

package main

import (
	"crypto/rc4"
	"fmt"
)

func main() {
	rc, err := rc4.NewCipher([]byte("5813aecc-0d05-497e-aafa-3111610cf44c"))

	if err != nil {
		panic(err)
	}
    
	src := []byte("hello world")
	fmt.Println(src)

	rc.XORKeyStream(src, src)
	fmt.Println(src)

	rc.XORKeyStream(src, src)
	fmt.Println(src)
}

输出：

[104 101 108 108 111 32 119 111 114 108 100]
[216 36 143 56 20 50 169 207 31 252 154]
[0 32 70 107 105 65 236 28 246 248 21]

Answer 1

看了下NewCipher：

// NewCipher creates and returns a new Cipher.  The key argument should be the
// RC4 key, at least 1 byte and at most 256 bytes.
func NewCipher(key []byte) (*Cipher, error) {
	k := len(key)
	if k < 1 || k > 256 {
		return nil, KeySizeError(k)
	}
	var c Cipher
	for i := 0; i < 256; i++ {
		c.s[i] = uint8(i)
	}
	var j uint8 = 0
	for i := 0; i < 256; i++ {
		j += c.s[i] + key[i%k]
		c.s[i], c.s[j] = c.s[j], c.s[i]
	}
	return &c, nil
}

XORKeyStream：

// XORKeyStream sets dst to the result of XORing src with the key stream.
// Dst and src may be the same slice but otherwise should not overlap.
func (c *Cipher) XORKeyStream(dst, src []byte) {
    for i := range src {
    	c.i += 1
    	c.j += c.s[c.i]
    	c.s[c.i], c.s[c.j] = c.s[c.j], c.s[c.i]
    	dst[i] = src[i] ^ c.s[c.s[c.i]+c.s[c.j]]
    }
}

这里并不是简单的异或运算，所以执行两次以后不会有xor运算的性质