Re: How to convert a string containing two hex digits to a character?

["Hibou57 (Yannick Duchêne)" <yannick_duchene@yahoo.fr>]

You may first convert from two 4 bits digits, that is, convert "7" and
"C".

16#7#; -- 7
16#C#; -- 12

To get 16#7C# from 16#7# and 16#C#, just do ((16#7# * 16) + 16#C#)

To get 16#7# and 16#C#, first get '7' and 'C' from the string, you may
do something like the following (providing your string is named S, and
it is an ASCII string) :

   High_Order_Digit_Position : constant Positive := S'First;
   Low_Order_Digit_Position : constant Positive := Positive'Succ
(High_Order_Digit_Position);
   High_Order_Digit_Representation : constant Character := S
(High_Order_Digit_Position);
   Low_Order_Digit_Representation : constant Character := S
(Low_Order_Digit_Position);

To get a digit Natural value from its Character representation, you
may do something like the following (providing your character is
stored in a Character named C) :

   case C is
      when '0' .. '9' =>
         Digit_Value := 0 + (Character'Pos (C) - Character'Pos ('0'));
      when 'A' .. 'F' =>
         Digit_Value := 10 + (Character'Pos (C) - Character'Pos
('A'));
      when 'a' .. 'f' =>
         Digit_Value := 10 + (Character'Pos (C) - Character'Pos
('a'));
      when others =>
         raise Program_Error;
   end case;

Wrap the latter in a function like “ function Hexadecimal_Digit_Value
(C : Character) return Natural; ”

Then do :

   Hexadecimal_Base : constant Natural := 16;
   High_Order_Digit : constant Natural := Hexadecimal_Digit_Value
(High_Order_Digit_Representation);
   Low_Order_Digit : constant Natural := Hexadecimal_Digit_Value
(Low_Order_Digit_Representation);
   Number : constant Natural := (High_Order_Digit * Hexadecimal_Base)
+ Low_Order_Digit;