Addition of 2 consts to var: how it's encoded?

[AlexTP]

Code: Pascal  [Select][+][-]

1.   function ParseHexDigit(c: char): Byte;
2.   begin
3.     case c of
4.       '0'..'9': Result := Byte(c) - Byte('0');
5.       'a'..'f': Result := Byte(c) - Byte('a') + 10;
6.       'A'..'F': Result := Byte(c) - Byte('A') + 10;
7.       else TryStringToGUID := False;
8.     end;
9.   end;
10.  

In this line
Result := Byte(c) - Byte('A') + 10
is this encoded as subtraction and then addition? Or as only one addition?

[ALLIGATOR]

O0: sub + add
O1,2,3: sub OR lea

[jamie]

StrToInt('$'+C)

[creaothceann]

Compiler Explorer to the rescue:

https://godbolt.org/z/Ynsb8ovjv

(ParseHexDigit_2 is a table look-up and should be much faster, since it has fewer instructions and is branch-free.)

Code: ASM  [Select][+][-]

1. parsehexdigit(char):
2.         movb    %dil,%al
3.         cmpb    $48,%al
4.         jb      .Lj6
5.         subb    $57,%al
6.         jbe     .Lj7
7.         subb    $8,%al
8.         jb      .Lj6
9.         subb    $5,%al
10.         jbe     .Lj9
11.         subb    $27,%al
12.         jb      .Lj6
13.         subb    $5,%al
14.         jbe     .Lj8
15.         jmp     .Lj6
16. .Lj7:
17.         movzbl  %dil,%eax
18.         subl    $48,%eax
19.         ret
20. .Lj8:
21.         movzbl  %dil,%edx
22.         subl    $97,%edx
23.         addl    $10,%edx
24.         movl    %edx,%eax
25.         ret
26. .Lj9:
27.         andl    $255,%edi
28.         subl    $65,%edi
29.         leal    10(%edi),%eax
30.         ret
31. .Lj6:
32.         xorl    %eax,%eax
33.         movq    trystringtoguid@GOTPCREL(%rip),%rdx
34.         movb    $0,(%rdx)
35.         ret

Code: ASM  [Select][+][-]

1. parsehexdigit_2(char):
2.         andl    $255,%edi
3.         leaq    tc_$output$_$.lut()(%rip),%rax
4.         movzbl  (%rax,%rdi,1),%eax
5.         cmpl    $128,%eax
6.         movq    trystringtoguid@GOTPCREL(%rip),%rdx
7.         setgeb  (%rdx)
8.         andl    $15,%eax
9.         ret

Code: ASM  [Select][+][-]

1. trystringtoguid:
2.         .zero 1
3.  
4. tc_$output$_$.lut():
5.         .byte   0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,128,129,130,131,132,133,134,135
6.         .byte   136,137,0,0,0,0,0,0,0,138,139,140,141,142,143,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
7.         .byte   0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
8.         .byte   0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0
9.         .byte   0,0,0

[jamie]

The compiler would be broken if it went out and first added 10 to Byte('A') and used the sum to subtract from BYTE(C)

Only / or * will do that and related multiplier operators.

Now if we are talking C language, there are some screwy things about that, but Pascal is supposed to follow what you learned in school.
Now, there is always the possibility the compiler could be optimizing those two on the right into a single value since they are constants, that would be a problem.
Jamie

[jamie]

Code: Pascal  [Select][+][-]

1. Function GetHex(C:Char):Integer; inline; //Returns -1 if not valid.
2. Const H:Array[0..15] of Char = '0123456789abcdef';
3. Begin
4.  Exit(IndexChar(H,SizeOF(H),Char(Byte(C) Or $20)));
5. end;                                                        
6.  

That will return -1 if not a vailed digit and also it in lines making only one call to the indexChar which is an RTL level call which should make things faster.

[creaothceann]

there is always the possibility the compiler could be optimizing those two on the right into a single value since they are constants, that would be a problem

That is exactly what the compiler is doing.

[jamie]

Looking at you ASM dump, I don't see that, it looks like it's adding 10 at the last step after it has already subtracted prior to.

P.S.

 I just ran his code here on 3.2.2 32 bits, it works ok and I did it inline, also.

jamie

[creaothceann]

It combines the operations after changing the code to this:

Code: Pascal  [Select][+][-]

1.                 '0'..'9': Result := ord(c) + (   - ord('0'));
2.                 'a'..'f': Result := ord(c) + (10 - ord('a'));
3.                 'A'..'F': Result := ord(c) + (10 - ord('A'));