Difference between adding a unit to the interface or the implementation section

Question

If I have a unit that is filled with constants like...

unit AConsts;
interface
const
   Const1 : WideString = \'Const1\';
   Const2 : WideString = \'Const2\'         


        

Answer 1

The difference has to do with where you're allowed to refer to the things that AConsts has in its interface section. In the first AUnit, you could use Const4 to declare a fixed-size array in that interface section. You couldn't do that in the second AUnit because Const4 isn't in scope.

It can have an effect on the compiled program, if you're not careful. Suppose we have another unit that also declares a constant named Const4:

unit BConsts;
interface
const
  Const4 = 50;
implementation
end.

Now we define an array in UnitA like this:

unit AUnit
interface
uses BConsts;
var
  data: array[0..Pred(Const4)] of Integer;
implementation
uses AConsts;
procedure Work;
var
  i: Integer;
begin
  for i := 0 to Const4 - 1 do begin
    data[i] := 8;
  end;
end;
end.

That code will write beyond the end of the array because the Const4 that's in scope in the interface section is not the same Const4 that's used in the implementation section. This doesn't happen often with constants. It usually just happens with two identifiers, the FindClose function defined in Windows and SysUtils, and TBitmap, defined in Graphics and Windows. And in those two cases, the compiler will tell you that you've done something wrong, although it won't tell you precisely that you've used an identifier that has two different meanings. You can resolve the problem by qualifying the identifier:

for i := 0 to BConsts.Const4 - 1 do
  data[i] := 8;

If all the above precautions are addressed, so your program compiles and runs correctly, then it makes no difference where units are used. In your example with App1 and App2, the two programs will be the same. They won't be identical — the compiler will have processed things in a different order and thus will likely put things in different places — but it will have no effect on the execution of your program.