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Subject: Re: How to convert a string containing two hex digits to a character?
Date: Tue, 5 Jan 2010 17:22:01 -0800 (PST)
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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 :=3D S'First;
   Low_Order_Digit_Position : constant Positive :=3D Positive'Succ
(High_Order_Digit_Position);
   High_Order_Digit_Representation : constant Character :=3D S
(High_Order_Digit_Position);
   Low_Order_Digit_Representation : constant Character :=3D 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' =3D>
         Digit_Value :=3D 0 + (Character'Pos (C) - Character'Pos ('0'));
      when 'A' .. 'F' =3D>
         Digit_Value :=3D 10 + (Character'Pos (C) - Character'Pos
('A'));
      when 'a' .. 'f' =3D>
         Digit_Value :=3D 10 + (Character'Pos (C) - Character'Pos
('a'));
      when others =3D>
         raise Program_Error;
   end case;

Wrap the latter in a function like =93 function Hexadecimal_Digit_Value
(C : Character) return Natural; =94

Then do :

   Hexadecimal_Base : constant Natural :=3D 16;
   High_Order_Digit : constant Natural :=3D Hexadecimal_Digit_Value
(High_Order_Digit_Representation);
   Low_Order_Digit : constant Natural :=3D Hexadecimal_Digit_Value
(Low_Order_Digit_Representation);
   Number : constant Natural :=3D (High_Order_Digit * Hexadecimal_Base)
+ Low_Order_Digit;