Converting Hex string to binary string

[munair]

I would take the opportunity and make three functions, which should limit moving around strings. Plus you get two extra conversion routines for free. I know the system lib provides it all, but its fun and educational. If I find time I will write the functions tomorrow.

[munair]

As promised.

Code: Pascal  [Select][+][-]

1. {$mode objFPC}
2.  
3. program hex2bin;
4.  
5. function dec2bin(d: longword): string;
6.   var
7.     n: byte = 65;
8. begin
9.   setlength(result, 64);
10.   if d = 0 then
11.     begin
12.       dec(n);
13.       result[n] := '0';
14.     end;
15.   while d > 0 do
16.     begin
17.       dec(n);
18.       result[n] := chr(d mod 2 + 48);
19.       d := d div 2;
20.     end;
21.   result := copy(result, n, 65 - n);
22. end;
23.  
24. function hex2dec(const h: string): longword;
25. var
26.   b: byte;
27.   i: word;
28. begin
29.   result := 0;
30.   for i := 1 to length(h) do
31.     begin
32.       b := ord(h[i]);
33.       case b of
34.         97..102:          { a..f = 10..15}
35.           b := b - 87;
36.         65..70:            { A..F = 10..15 }
37.           b := b - 55;
38.         48..57:
39.           b := b - 48;
40.       end;
41.       result := result * 16 + b;
42.     end;
43. end;
44.  
45. function hex2bin(h: string): string;
46. begin
47.   result := dec2bin(hex2dec(h));
48. end;
49.  
50.  
51. begin
52.   //writeln(hex2dec('3b7'));
53.   //writeln(dec2bin(951));
54.   writeln(hex2bin('3b7'));
55. end.
56.  

[engkin]

Needed a break from what I was working on. Here is the result:

Code: Pascal  [Select][+][-]

1. function H2B(const h:string):string;
2. const
3.   bins:array[$0..$F] of string =('0000','0001','0010','0011','0100','0101','0110','0111','1000','1001','1010','1011','1100','1101','1110','1111');
4.  
5. type
6.   TNibbles=bitpacked record
7.     case boolean of
8.       False:(c:char);
9.       True: (L:$0..$F; H:$0..$F)
10.   end;
11.  
12. var
13.   b:PDWord;
14.   n:TNibbles;
15.   i:Integer;
16. begin
17.   SetLength(Result,Length(h) shl 2);
18.   b:=@Result[1];
19.   for i:=1 to Length(h) do
20.   begin
21.     n.c:=h[i];
22.     case n.H of
23.       3:  begin if n.L < 10      then Move(bins[0+n.L][1],b^,4) else exit('Error') end;
24.       4,6:begin if n.L in [1..6] then Move(bins[9+n.L][1],b^,4) else exit('Error') end;
25.       else;
26.         Exit('Error');
27.     end;
28.     inc(b);
29.   end;
30. end;

[engkin]

@jamie, looks nice, but it doesn't seem to produce binary string.

[Kays]

[…] One item I can't remember is the clause used to inform the compiler registers […]

This

Code: Pascal  [Select][+][-]

1. end ['rax', 'rcx'];

is only necessary for certain registers in pure assembly routines like yours.

And if you’re already writing assembly, you definitely want to process (at least) eight Bytes, eight chars at once, because the compiler generated code using Pascal source code definitely beats your implementation.

[engkin]

Jamie, your code does not give "binary string".

It does not, for instance, produce '1110' when given 'E'

Your code is Hex String to Decimal, or did I miss something?

[440bx]

I don't understand here.

The idea was to take a HEX string of chars and generate a binary of it.

a binary string, i.e, a string of '0's and '1's - not the binary number represented by a hex string.

[Seenkao]

Needed a break from what I was working on. Here is the result:

Code: Pascal  [Select][+][-]

1. function H2B(const h:string):string;
2. const
3.   bins:array[$0..$F] of string =('0000','0001','0010','0011','0100','0101','0110','0111','1000','1001','1010','1011','1100','1101','1110','1111');
4.  
5. type
6.   TNibbles=bitpacked record
7.     case boolean of
8.       False:(c:char);
9.       True: (L:$0..$F; H:$0..$F)
10.   end;
11.  
12. var
13.   b:PDWord;
14.   n:TNibbles;
15.   i:Integer;
16. begin
17.   SetLength(Result,Length(h) shl 2);
18.   b:=@Result[1];
19.   for i:=1 to Length(h) do
20.   begin
21.     n.c:=h[i];
22.     case n.H of
23.       3:  begin if n.L < 10      then Move(bins[0+n.L][1],b^,4) else exit('Error') end;
24.       4,6:begin if n.L in [1..6] then Move(bins[9+n.L][1],b^,4) else exit('Error') end;
25.       else;
26.         Exit('Error');
27.     end;
28.     inc(b);
29.   end;
30. end;

Можно немного проще
Eng: Can be a little simpler 

Code: Pascal  [Select][+][-]

1. var
2.   bb: integer;
3.   ...
4. for i := 1 to Length(h) do
5.   begin
6.     bb := Byte(h[i]);
7.     case bb of
8.         48..57: Result := Result + bins[bb - 48];
9.         65..70: Result := Result + bins[bb - 55];
10.         97..102: Result := Result + bins[bb - 87];
11.       else begin
12.         Result := 'Error';
13.         exit;
14.       end;
15.     end;
16.   end;      

Иногда простые вещи могут быть более эффективными и понятными. Но ваше решение мне понравилось!
Eng: Sometimes simple things can be more effective and understandable. But I liked your decision! 

[engkin]

Yes, definitely.