无符号字符到char *和C中的int？

Question

Working on some encryption that requires unsigned char's in the functions, but want to convert to a char for use after it's been decrypted. So, I have:

unsigned char plaintext[16];
char *plainchar;
int plainint;

... Code now populates plaintext with data that happens to all be plain text

Now at this point, let's say plaintext is actually a data string of "0123456789". How can I get the value of plaintext into plainchar as "012456789", and at the same time plainint as 123456789?

-- Edit --

Doing this when plaintext is equal to "AAAAAAAAAA105450":

unsigned char plaintext[16];
char *plainchar;
int plainint;

... Code now populates plaintext with data that happens to all be plain text

plainchar = (char*)plaintext;

Makes plainchar equal to "AAAAAAAAAA105450╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠╠┤∙7" with a sizeof = 51. The encryption code is the rijndael example code, so it should be working fine.

Thanks, Ben

Answer 1

Your plain text string is not terminated. All strings must have an extra character that tells the end of the string. This character is '\\0' .

Declare the plaintext variable as

unsigned char plaintext[17];

and after you are done with the decryption add this

plaintext[last_pos] = '\0';

Change last_pos to the last position of the decrypted text, default to 16 (last index of the array).

Answer 2

I think its simply

plainchar = (char*)plaintext;
sscanf( plainchar, "%d", &plainint );

Answer 3

for unsigned char to char*

plainchar = (char*)plaintext;

for unsigned to int

sscanf( plainchar, "%d", &plainint );