Hi friends, I was working on aprogram that takes 16 Hex characters and converts to 8-ascii characters the program generates large numbers which are in bytes, but somehow I cannot change to correponding Ascii characters. it generates numbers such 166,239... numbers less than 255. the function calls Convert() which returns the decimal of hex character Can anyone help me how I can achieve the desired Ascii chars ? the code is commented to be able to read. I am really in need for help to get this wright. Thanks in advance. /* prtotypes */ typedef unsigned char ByteType ; int Convert(ByteType val ); ByteType ASCIZ(ByteType *Addr); ByteType ASCIZ(ByteType *Addr) { int i, v1, v2; ByteType a1, a2, v,y; /* Convert 16-HEXA-character key to be 8-ASCII-character key. */ for (i = 0; i < 8; i++) { a1 = *Addr++; if (i == 7) a2 = *Addr; else a2 = *Addr++; /* Convert two HEXA characters into their decimal value. */ v1 = Convert(a1); v2 = Convert(a2); /* Convert the two decimal value to be one ASCII character. */ v = ((v1 * 16) + v2); } return v; }

The ASCII range is from 0 to 127, of which 0 to 31 and 127 are non printing characters, well sort of. Problem 1: how are you intending to represent a character over 128 unless you go for an extended character set. I don't know if there is a standard extension. Problem 2: if you are printing them, how are you going to print the non-printable characters. Are you printing them? You could go for the uuencode type strategy which restricts the set. The result of this is that you will need more than 8 chars for 16 digits.