我知道理论上河豚比aes快得多。但是,我在java 8平台和弹跳城堡库中对一些算法进行了测试,包括aes和用于1MB、5MB、10MB等文件的河豚。在每一种测试场景中,aes都比河豚快。
我想知道我在哪里犯了错误?
以下是代码:
private static final int WARMUP_COUNT = 5;
private static final int FILE_LENGTH = 1024*512;
private static final int ITERATOR_COUNT = 1000;
private static final double BOLME = 1_000_000.0 * (ITERATOR_COUNT-WARMUP_COUNT);
static final private byte[] ivBytes = new byte[] { 0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00};
private static final IvParameterSpec ivSpec16bytes = new IvParameterSpec(ivBytes);
private static final IvParameterSpec ivSpec8bytes = new IvParameterSpec(Arrays.copyOfRange(ivBytes,0,8));
static String[] algosWithMode = {"AES/CBC/PKCS7Padding","Blowfish/CBC/PKCS7Padding","CAST5/CBC/PKCS7Padding","DES/CBC/PKCS7Padding","DESede/CBC/PKCS7Padding", "IDEA/CBC/PKCS7Padding","ARC4", };
static String[] algos = { "AES","Blowfish","CAST5","DES", "DESede","IDEA","ARC4" };
static int[] keylenngth = {128,128,128,56, 168,128,128 };
@SuppressWarnings("unused")
public static void main(String[] args) throws Exception {
if(ITERATOR_COUNT <= WARMUP_COUNT )
throw new Exception("iterator count must be greater than warm up count iterator: "+ITERATOR_COUNT
+" warmup count :" + WARMUP_COUNT);
Security.addProvider(new BouncyCastleProvider());
Key key = null;
byte[] plainText=null;
byte[] cipherText=null;
byte[] decryptedText=null;
long startTime;
DecimalFormat df = new DecimalFormat("0.000");
for (int k = 0; k < 7; k++) {
long timeDec = 0,timeEnc = 0,timekey = 0;
long maxtimeDec = 0,maxtimeEnc = 0,maxtimekey = 0;
long mintimeDec = Long.MAX_VALUE,mintimeEnc = Long.MAX_VALUE,mintimekey = Long.MAX_VALUE;
long topDec = 0,topEnc = 0,topkey = 0;
for (int i = 0; i < ITERATOR_COUNT; i++) {
SecureRandom random= new SecureRandom();
plainText = random.generateSeed(FILE_LENGTH);
startTime=System.nanoTime();
KeyGenerator keyGen = KeyGenerator.getInstance(algos[k]);
keyGen.init(keylenngth[k],random);
key=keyGen.generateKey();
timekey=System.nanoTime()-startTime;
Cipher cipher=null;
if(k == 0){
cipher = Cipher.getInstance(algosWithMode[k]);
cipher.init(Cipher.ENCRYPT_MODE, key,ivSpec16bytes);
}else if(k == 6){
cipher = Cipher.getInstance(algosWithMode[k]);
cipher.init(Cipher.ENCRYPT_MODE, key);
}else{
cipher = Cipher.getInstance(algosWithMode[k]);
cipher.init(Cipher.ENCRYPT_MODE, key,ivSpec8bytes);
}
startTime=System.nanoTime();
cipherText = cipher.doFinal(plainText);
timeEnc=System.nanoTime()-startTime;
if(k == 0){
cipher = Cipher.getInstance(algosWithMode[k]);
cipher.init(Cipher.DECRYPT_MODE, key,ivSpec16bytes);
}else if(k== 6){
cipher = Cipher.getInstance(algosWithMode[k]);
cipher.init(Cipher.DECRYPT_MODE, key);
}else {
cipher = Cipher.getInstance(algosWithMode[k]);
cipher.init(Cipher.DECRYPT_MODE, key,ivSpec8bytes);
}
startTime=System.nanoTime();
cipher.doFinal(cipherText);
timeDec=System.nanoTime()-startTime;
if (i >= WARMUP_COUNT) {
if (maxtimeEnc < timeEnc)
maxtimeEnc = timeEnc;
if (maxtimeDec < timeDec)
maxtimeDec = timeDec;
if (maxtimekey < timekey)
maxtimekey = timekey;
if (mintimeEnc > timeEnc)
mintimeEnc = timeEnc;
if (mintimeDec > timeDec)
mintimeDec = timeDec;
if (mintimekey > timekey)
mintimekey = timekey;
topEnc += timeEnc;
topDec += timeDec;
topkey += timekey;
}
}
double avgEnc=topEnc/BOLME;
double avgDec=topDec/BOLME;
double avgKey=topkey/BOLME;
System.out.println("********************************************************"+algos[k]+"*****************************************************************");
System.out.println("Avg Enc :"+df.format(avgEnc)+" - "+" Avg Dec :"+df.format(avgDec)+"-"+" Avg Key :"+ df.format(avgKey));
System.out.println("Max Enc :"+df.format(maxtimeEnc/1_000_000.0)+" - "+" Max Dec :"+df.format(maxtimeDec/1_000_000.0)+"-"+" Max Key :"+ df.format(maxtimekey/1_000_000.0));
System.out.println("Min Enc :"+df.format(mintimeEnc/1_000_000.0)+" - "+" Min Dec :"+df.format(mintimeDec/1_000_000.0)+"-"+" Min Key :"+ df.format(mintimekey/1_000_000.0));
System.out.println();
//System.out.println();
}
}发布于 2014-05-26 15:30:38
理论上,Blowfish应该比AES更快,但不会更快。有关详细信息,请参阅这个问题。
然后是优化。对于给定的算法,您可以以某种方式将其“最高速度”定义为使用最优代码实现的性能;但是实际的实现从来都不是完全最优的,它们离最优性有多近(或多远)取决于它们所投入的努力。AES很大,而且使用很多;所以编写AES实现的人要小心地调优他们的代码,以便在大多数体系结构(包括最新的架构)上快速运行。另一方面,河豚有一个越来越少的使用基础,有很好的理由(它有64位块,不足以确保安全时,当千兆字节的数据是用一个给定的密钥加密)。因此,可以假设在给定的密码库中,AES实现比Blowfish实现更有可能得到优化和维护。
其他要点:
发布于 2014-05-26 15:14:26
现代的x86 CPU为AES加密/解密提供了硬件加速,即在JVM中启用。虽然在软件实现中,Blowfish在AES中可能更快,但是硬件加速使它更快。
顺便说一下,这个问题应该是堆栈溢出的,而不是这里。
https://security.stackexchange.com/questions/58756
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