Reputation: 184
Sha256 hash function gives a longer hashed string in objective c than Java. Extra Zeros being added in objective C, how can I rationalise the hashing?
Objective C:
-(NSString*) sha256:(NSString *)clear{
const char *s=[clear cStringUsingEncoding:NSASCIIStringEncoding];
NSData *keyData=[NSData dataWithBytes:s length:strlen(s)];
uint8_t digest[CC_SHA256_DIGEST_LENGTH]={0};
CC_SHA256(keyData.bytes, keyData.length, digest);
NSData *out=[NSData dataWithBytes:digest
length:CC_SHA256_DIGEST_LENGTH];
NSString *hash=[out description];
hash = [hash stringByReplacingOccurrencesOfString:@" " withString:@""];
hash = [hash stringByReplacingOccurrencesOfString:@"<" withString:@""];
hash = [hash stringByReplacingOccurrencesOfString:@">" withString:@""];
return hash;
}
Java
public static String generateHashString(String data)
{
try {
MessageDigest md = MessageDigest.getInstance("SHA-256");
byte[] dataInBytes = data.getBytes(StandardCharsets.UTF_8);
md.update(dataInBytes);
byte[] mdbytes = md.digest();
StringBuffer hexString = new StringBuffer();
for (int i=0;i<mdbytes.length;i++) {
hexString.append(Integer.toHexString(0xFF & mdbytes[i]));
}
return hexString.toString();
} catch (NoSuchAlgorithmException e) {
e.printStackTrace();
}
return null;
}
Upvotes: 3
Views: 1235
Reputation: 39218
Integer.toHexString() on an integer less than 16 will only be one character long, whereas you want the extra '0' character.
You could use String.format():
for (int i = 0; i < mdbytes.length; i++) {
hexString.append(String.format("%02x", 0xFF & mdbytes[i]));
}
Also, you really should be using StringBuilder
rather than StringBuffer
in this case because only a single thread is involved.
See Java code To convert byte to Hexadecimal for some alternative solutions to hex-encoding a byte array in Java.
Upvotes: 1