Alex Rivera | Logout

How to SHA1 hash a string in Android?

Asked 2011-05-12T15:42:24.277
77

In Objective C I've been using the following code to hash a string:

-(NSString *) sha1:(NSString*)stringToHash {    
    const char *cStr = [stringToHash UTF8String];
    unsigned char result[20];
    CC_SHA1( cStr, strlen(cStr), result );
    return [NSString stringWithFormat:@"%02X%02X%02X%02X%02X%02X%02X%02X%02X%02X%02X%02X%02X%02X%02X%02X%02X%02X%02X%02X",
        result[0], result[1], result[2], result[3], 
        result[4], result[5], result[6], result[7],
        result[8], result[9], result[10], result[11],
        result[12], result[13], result[14], result[15],
        result[16], result[17], result[18], result[19]
        ];  
}

Now I need the same for Android but can't find out how to do it. I've been looking for example at this: Make SHA1 encryption on Android? but that doesn't give me the same result as on iPhone. Can anyone point me in the right direction?

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1 Answer

39

A simpler SHA-1 method: (updated from the commenter's suggestions, also using a massively more efficient byte->string algorithm)

String sha1Hash( String toHash )
{
    String hash = null;
    try
    {
        MessageDigest digest = MessageDigest.getInstance( "SHA-1" );
        byte[] bytes = toHash.getBytes("UTF-8");
        digest.update(bytes, 0, bytes.length);
        bytes = digest.digest();

        // This is ~55x faster than looping and String.formating()
        hash = bytesToHex( bytes );
    }
    catch( NoSuchAlgorithmException e )
    {
        e.printStackTrace();
    }
    catch( UnsupportedEncodingException e )
    {
        e.printStackTrace();
    }
    return hash;
}

// http://stackoverflow.com/questions/9655181/convert-from-byte-array-to-hex-string-in-java
final protected static char[] hexArray = "0123456789ABCDEF".toCharArray();
public static String bytesToHex( byte[] bytes )
{
    char[] hexChars = new char[ bytes.length * 2 ];
    for( int j = 0; j < bytes.length; j++ )
    {
        int v = bytes[ j ] & 0xFF;
        hexChars[ j * 2 ] = hexArray[ v >>> 4 ];
        hexChars[ j * 2 + 1 ] = hexArray[ v & 0x0F ];
    }
    return new String( hexChars );
}
answered 2012-08-16T00:00:51.777

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