Anyone know the solution of that problem? Exited with code 216

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hello, this is my program about lifo animation, there isn't syntax error, but when i want try to menu number 2, there is exit code 216. anyone can help me? thank you.

program SLL_LIFO;
uses crt;
type
        pointermhs = ^recmhs;
        recmhs = record
        huruf : char;
        berikut : pointermhs;
end;

var kepala, ekor, isi : pointermhs;
pil : integer;
i   : longint;
n   : byte;

procedure buat_simpul_lifo;
begin
        new(isi);
        if kepala=nil then
                isi^.berikut := nil
        else
                isi^.berikut := kepala;
        kepala:=isi;
end;

procedure menu;
begin
gotoxy(1,1);

        writeln('MENU PROGRAM');
        writeln('1. Masuk');
        writeln('2. Keluar');
        gotoxy(1,4); writeln('3. Selesai');
        gotoxy(17,12); writeln('===============================');
        gotoxy(17,14); writeln('===============================');

        write('pilih nomor :');readln(pil);

end;

procedure Entry;
var i :byte;
begin

        inc(n);
        buat_simpul_lifo;
        write('huruf :');readln(isi^.huruf);
        for i := 8 to 47-n*3 do
        begin
                delay(30);
                gotoxy(i,13);write('  ',isi^.huruf);
        end;
        readkey;
end;

procedure hapus;
var i:longint;
n:byte;
begin

        for i:= 47-n*3 downto 8 do
        begin
                delay(30);
                gotoxy(i,13);write('  ');write(isi^.huruf);
        end;
        dec(n);
        dispose(isi);
        isi:=kepala;
        readkey;

end;


begin
clrscr;
n:=0;

                repeat
                {clrscr;}
                menu;
                kepala:=nil;
                ekor:=nil;
                CASE PIL OF
                        1 : Entry;
                        2 : hapus;
                        3 : exit;


                END;
        until pil =4;

END.

What I have tried:

program SLL_LIFO;
uses crt;
type
        pointermhs = ^recmhs;
        recmhs = record
        huruf : char;
        berikut : pointermhs;
end;

var kepala, ekor, isi : pointermhs;
pil : integer;
i   : longint;
n   : byte;

procedure buat_simpul_lifo;
begin
        new(isi);
        if kepala=nil then
                isi^.berikut := nil
        else
                isi^.berikut := kepala;
        kepala:=isi;
end;

procedure menu;
begin
gotoxy(1,1);

        writeln('MENU PROGRAM');
        writeln('1. Masuk');
        writeln('2. Keluar');
        gotoxy(1,4); writeln('3. Selesai');
        gotoxy(17,12); writeln('===============================');
        gotoxy(17,14); writeln('===============================');

        write('pilih nomor :');readln(pil);

end;

procedure Entry;
var i :byte;
begin

        inc(n);
        buat_simpul_lifo;
        write('huruf :');readln(isi^.huruf);
        for i := 8 to 47-n*3 do
        begin
                delay(30);
                gotoxy(i,13);write('  ',isi^.huruf);
        end;
        readkey;
end;

procedure hapus;
var i:longint;
n:byte;
begin

        for i:= 47-n*3 downto 8 do
        begin
                delay(30);
                gotoxy(i,13);write('  ');write(isi^.huruf);
        end;
        dec(n);
        dispose(isi);
        isi:=kepala;
        readkey;

end;


begin
clrscr;
n:=0;

                repeat
                {clrscr;}
                menu;
                kepala:=nil;
                ekor:=nil;
                CASE PIL OF
                        1 : Entry;
                        2 : hapus;
                        3 : exit;


                END;
        until pil =4;

END.

Posted 24-Feb-20 4:10am

Member 14753870

Updated 24-Feb-20 4:31am

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OriginalGriff · Answer 1 · 2020-02-24T04:24:00

Compiling does not mean your code is right! :laugh:
Think of the development process as writing an email: compiling successfully means that you wrote the email in the right language - English, rather than German for example - not that the email contained the message you wanted to send.

So now you enter the second stage of development (in reality it's the fourth or fifth, but you'll come to the earlier stages later): Testing and Debugging.

Start by looking at what it does do, and how that differs from what you wanted. This is important, because it give you information as to why it's doing it. For example, if a program is intended to let the user enter a number and it doubles it and prints the answer, then if the input / output was like this:

Input   Expected output    Actual output
  1            2                 1
  2            4                 4
  3            6                 9
  4            8                16

Then it's fairly obvious that the problem is with the bit which doubles it - it's not adding itself to itself, or multiplying it by 2, it's multiplying it by itself and returning the square of the input.
So with that, you can look at the code and it's obvious that it's somewhere here:

C#

int Double(int value)
   {
   return value * value;
   }

Once you have an idea what might be going wrong, start using the debugger to find out why. Put a breakpoint on the first line of the method, and run your app. When it reaches the breakpoint, the debugger will stop, and hand control over to you. You can now run your code line-by-line (called "single stepping") and look at (or even change) variable contents as necessary (heck, you can even change the code and try again if you need to).
Think about what each line in the code should do before you execute it, and compare that to what it actually did when you use the "Step over" button to execute each line in turn. Did it do what you expect? If so, move on to the next line.
If not, why not? How does it differ?
Hopefully, that should help you locate which part of that code has a problem, and what the problem is.
This is a skill, and it's one which is well worth developing as it helps you in the real world as well as in development. And like all skills, it only improves by use!

Richard MacCutchan · Answer 2 · 2020-02-24T04:32:00

Solution 2

You do not have any explicit return value in your main code so the exit code will be some random value: e.g. 216. Try changing option 3 from

Delphi

3: exit;

to

Delphi

3: Exit(0);

Posted 24-Feb-20 4:32am

Richard MacCutchan