Compiler ignores missing parenthesis of Exit command

Question

Why does the Delphi compiler ignore this missing parenthesis?

function Test: Boolean;
begin
  Exit(True;    // <-- eek! it compiles...
end;

I found some of my code looking like this and first thought that Delphi ignores my unit - but it just ignores this type of syntax error. So now of course I want to know why.

my guess it would be that since the function contains nothing else , it ignores the function itself. Try adding some code before the exit . — CyprUS, Apr 30 '12 at 08:15
@CyprUS The context it came from was more complex. But I also played with it a bit. E.g. removing the semicolon as well makes it complain again. — Heinrich Ulbricht, Apr 30 '12 at 08:27
in delphi 2006 gives the error "E2066 Missing operator or semicolon", probably a bug in xe2 — RBA, Apr 30 '12 at 08:40
@RBA The parameter for Exit was introduced in Delphi 2009 IIRC — Heinrich Ulbricht, Apr 30 '12 at 08:43
Does not compile under Free Pascal 2.7.1, even in Delphi mode, FWIW (says **Fatal: Syntax error, ")" expected but ";" found**), so it does look like a compiler bug. — Thomas, Apr 30 '12 at 09:06
eek! MkII. I didn't even know that Exit had a parameter now. — Martin James, Apr 30 '12 at 12:05

score 1 · Answer 1 · answered Apr 30 '12 at 08:17

1

I'm guessing Exit is considered a token unto itself, and as such anything defined within the same scope after Exit is simply ignored by the compiler (since it cannot execute those instructions anyway).

answered Apr 30 '12 at 08:17

LaKraven

5,804
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Adding instructions after the `Exit(True;` makes the compiler complain again with "E2066 Missing operator or semicolon" (Delphi XE). – Heinrich Ulbricht Apr 30 '12 at 08:30
I'd presume this is because the semicolon prompts the compiler to attempt to compile the next instruction. – LaKraven Apr 30 '12 at 08:32
Essentially I'm saying that `Exit` is an "absolute" token, prompting the compiler to disregard any subsequent chars on the same statement until the next occurence of the semicolon, which separates it from the proceeding instruction(s). It's a quirk, for sure... and an interesting find! – LaKraven Apr 30 '12 at 08:33

score 0 · Answer 2 · answered May 04 '12 at 12:55

0

Maybe the compiler is thinking that either 1. There is an Exit by itself, or 2. There is an Exit with a set of parentheses (). If it doesn't find #2 it goes to #1.

answered May 04 '12 at 12:55

Patrick Moloney

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Compiler ignores missing parenthesis of Exit command

2 Answers2