Can you please clarify?
Is it okay to pass as HEX encoded unicode? Does it strip out the zeros?
For example, if my "block" was 3 bytes of all AAA hex encoded is 414141.
Do you pass L"414141" which is actually 16-bytes "410041004100". If the IV was block size of 2 would it just use 4100->A or would it internally process 414141->AAA?
Hope this makes sense. Basically I'm asking if the AES block is 16 bytes:
Just wanted to verify.
asked Jul 20 at 10:59
Encryption algorithms operate on bytes, and therefore the byte representation of a string becomes very important. Chilkat solves this problem by understanding that in different programming languages, a "string" can be many different things. It can be an object. It can be implicitly utf-16. It can be implicitly ANSI or utf-8, etc.
The Chilkat solution is to provide a Crypt2.Charset property that indicates the desired byte representation to be used for encryption. This way, you don't have to worry about the string passed in -- whether it's from C# or if it's a "wchar_t *" from C++, etc. However, in cases where there is implicit ambiguity, such as a "const char *" in C/C++, then one must take care to tell the Chilkat API *what* is getting passed (ANSI or utf-8).
So... let's say you have the string "É". This is a single accented character found in Western European languages.
In the iso-8859-1 character encoding, it is represented by a single byte: 0xC9
To control what byte representation is actually passed to the internal encryptor, just set the Charset property equal to "ansi", "utf-8", "utf-16", or whatever. The full set of accepted charsets is listed here: http://cknotes.com/chilkat-charsets-character-encodings-supported/
Regarding block size -- that doesn't really have anything to do with the byte representation of characters. The block size is defined by the algorithm (for example AES encryption in ECB/CBC modes always pads to a multiple of the AES algorithm's block size, which is always 16 bytes (regardless of key size).