问题
Here is my code snippet:
int eValue = 79, t;
int bitLength = 1024; // KeySize
BigInteger e = new BigInteger(Integer.toString(eValue));
KeyPairGenerator kpg = KeyPairGenerator.getInstance("RSA");
kpg.initialize(bitLength);
KeyPair kp = kpg.generateKeyPair();
KeyFactory kfactory = KeyFactory.getInstance("RSA");
RSAPublicKeySpec kspec = (RSAPublicKeySpec) kfactory.getKeySpec(kp.getPublic(),
RSAPublicKeySpec.class);
System.out.println("Byte Length is : " + kspec.getModulus().toByteArray().length);
String testString;
try {
testString = new String (kspec.getModulus().toByteArray() , "ISO-8859-1");
StringBuilder tt = new StringBuilder();
for(t =0 ; t< testString.length() ; t++)
{
tt.append((int) testString.charAt(t)+",");
}
String encryptedBytes = tt.toString();
System.out.println("Mod is : " + encryptedBytes);
}catch (Exception ex) {
// TODO: handle exception
}
And here is the output:
Byte Length is : 129
Mod is : 0,190,193,141,230,128,124,6,201,254,135,66,162,65,147,160,76,160,181,7,141,113,8,57,193,185,206,42,125,9,169,209,124,74,233,151,10,128,180,35,24,206,213,32,48,4,39,178,60,10,249,151,50,218,220,11,124,72,64,148,135,251,133,23,54,171,25,202,157,28,21,39,239,234,48,56,79,36,127,59,203,108,189,232,216,231,237,237,90,253,19,118,29,18,142,126,254,193,189,82,15,126,139,136,45,31,133,242,187,81,62,52,5,23,11,217,171,233,7,137,115,30,93,206,236,31,196,111,153
For 1024 bit lenght key modulus should be 128 bytes and for 2048 it should be 256, but I am getting one byte extra (Adding 0 always at the very first byte), Need Help to reslove this..
Thanks, Pawan
回答1:
The reason for the 00h valued byte at the start is because BigInteger.toByteArray() returns the signed representation. As long as the key length in bits is N*8 (or key length % 8 = 0) then the signed representation of an RSA modulus will always feature a 00h valued byte at the start.
Simply remove the initial byte if it is zero by copying it into an array of the key length in bytes. Note that if you have a private exponent, it might also be shorter than the key length in bytes, so copy it to the end of the new byte array. Normally, this kind of method is known as I2OS or I2O (integer to octet string), where the octet sting (byte array in java) has a specified length.
/**
* Encodes the given value as a unsigned Big Endian within an octet string
* of octetStringSize bytes.
*
* @param i
* the integer to encode
* @param octetStringSize
* the number of octets in the octetString returned
* @return the encoding of i
* @throws IllegalArgumentException
* if the given integer i is negative
* @throws IllegalArgumentException
* if the octetStringSize is zero or lower
* @throws IllegalArgumentException
* if the given BigInteger does not fit into octetStringSize
* bytes
*/
public static byte[] integerToOctetString(final BigInteger i,
final int octetStringSize) {
// throws NullPointerException if i = null
if (i.signum() < 0) {
throw new IllegalArgumentException(
"argument i should not be negative");
}
if (octetStringSize <= 0) {
throw new IllegalArgumentException("octetStringSize argument ("
+ octetStringSize
+ ") should be higher than 0 to store any integer");
}
if (i.bitLength() > octetStringSize * Byte.SIZE) {
throw new IllegalArgumentException("argument i (" + i
+ ") does not fit into " + octetStringSize + " octets");
}
final byte[] signedEncoding = i.toByteArray();
final int signedEncodingLength = signedEncoding.length;
if (signedEncodingLength == octetStringSize) {
return signedEncoding;
}
final byte[] unsignedEncoding = new byte[octetStringSize];
if (signedEncoding[0] == (byte) 0x00) {
// skip first padding byte to create a (possitive) unsigned encoding for this number
System.arraycopy(signedEncoding, 1, unsignedEncoding,
octetStringSize - signedEncodingLength + 1,
signedEncodingLength - 1);
} else {
System.arraycopy(signedEncoding, 0, unsignedEncoding,
octetStringSize - signedEncodingLength,
signedEncodingLength);
}
return unsignedEncoding;
}
/**
* Returns a BigInteger that is the value represented by the unsigned, Big
* Endian encoding within the given octetString.
*
* @param octetString
* the octetString containing (only) the encoding
* @return the value represented by the octetString
*/
public static BigInteger octetStringToInteger(final byte[] octetString) {
// arguments are signum, magnitude as unsigned, Big Endian encoding
return new BigInteger(1, octetString);
}
/**
* Returns the minimum number of bytes required to directly store the given
* number of bits.
*
* @param bitSize
* the bitSize
* @return the size as a number of bytes
* @throws IllegalArgumentException
* if the given bitSize argument is negative
*/
public static int bitSizeToByteSize(final int bitSize) {
if (bitSize < 0) {
throw new IllegalArgumentException("bitSize (" + bitSize
+ " should not be negative");
}
return (bitSize + Byte.SIZE - 1) / Byte.SIZE;
}
回答2:
You can use Arrays.deepToString() to print a byte array directly:
String encryptedBytes = Arrays.deepToString(new Object[] { kspec.getModulus().toByteArray() })
I suspect you are having trouble with signed vs. unsigned numbers. The 128-bit modulus is unsigned, but to store it in a BigInteger might sometimes take 129 bits, hence the extra byte.
回答3:
As Maarten Bodewes has answered, the extra byte is space for the BigInteger's sign.
If the expected size is known and Hex is acceptable, i would use something like this:
System.out.printf("Mod is : %0256x%n" , kspec.getModulus());
来源：https://stackoverflow.com/questions/8515691/getting-1-byte-extra-in-the-modulus-rsa-key-and-sometimes-for-exponents-also