[SOLVED] TAChart hours displayed backwards

[lucamar]

This function (taken from this forum post) may be quicker, but I don't think it would be that much (unless there are thousands of calls to it):

Code: Pascal  [Select][+][-]

1. function Fun_IsLeapYear(aYear: integer): boolean;
2. begin
3.   Result := (aYear and 3) = 0;   // Test simple case
4.   if Result then begin
5.     aYear:=aYear shr 2;          // Reduce (Divide by 4)
6.     Result:=(aYear mod 25)<>0;   // Eliminate those that cannot
7.     if not Result then           // ..be divided by 100, they are always leaps
8.       Result:=(aYear and 3)=0;   // Those that can be divided by 4(00) are leaps
9.   end;
10. end;

If it has any advantage it may be that it can be easily re-implemented in optimized assembler.

[munair]

Quite revealing. With FPC year and 3 is more than twice as fast as year mod 4. With FBC it is the opposite. My benchmark shows that both LeapYear functions are equally fast. The third version proposed by Lucamar is slightly slower and has the disadvantage that it changes the function parameter.

Code: Pascal  [Select][+][-]

1. function aLeapYear1(const y: integer): boolean;
2. { http://forum.lazarus.freepascal.org/index.php/topic,43055.msg300956.html#msg300956 }
3. begin
4.   result := false;
5.   if (y and 3) = 0 then
6.     // divisible by 4
7.     begin
8.       if (y mod 100) = 0 then
9.         // divisible by 100
10.         begin
11.           // must also be divisible by 400
12.           if (y mod 400) = 0 then
13.             result := true
14.         end
15.       else
16.         result := true;
17.     end;
18. end;

Code: Pascal  [Select][+][-]

1. function aLeapYear2(const Year: integer): boolean;
2. { http://forum.lazarus.freepascal.org/index.php/topic,43055.msg300960.html#msg300960 }
3. begin
4.   Result := ((Year and 3) = 0) and ((Year mod 100 <> 0) or (Year mod 400 = 0));
5. end;

I will leave it at that. Further testing is better done in a new topic.