What happens when AnsiString is cast to PAnsiString?

Question

I have the method (Delphi 2009):

procedure TAnsiStringType.SetData(const Value: TBuffer; IsNull: boolean = False);
begin
  if not IsNull then
    FValue:= PAnsiString(Value)^;
  inherited;
end;

This is an abstract method on the base class, where "Value: Pointer" expects the pointer of the corresponding data, as:

String = PString
AnsiString = PAnsiString
Integer = PInteger
Boolean = PBoolean

So I try to pass the value like this:

var
  S: AnsiString;
begin
  S:= 'New AnsiString Buffer';
  SetBuffer(PAnsiString(S));
end;

But a cast from AnsiString to PAnsiString does NOT work, I can see why, but I want to know what the result of the casting is. So I wrote a simple test:

var
  Buffer: AnsiString;
  P1: Pointer;
  P2: Pointer;
  P3: Pointer;
  P4: Pointer;
begin
  P1:= PAnsiString(Buffer);
  P2:= Addr(Buffer);
  P3:= @Buffer;
  P4:= Pointer(Buffer);
  P5:= PChar(Buffer[1]);

  WriteLn('P1: ' + IntToStr(Integer(P1)));
  WriteLn('P2: ' + IntToStr(Integer(P2)));
  WriteLn('P3: ' + IntToStr(Integer(P3)));
  WriteLn('P4: ' + IntToStr(Integer(P4)));
  WriteLn('P5: ' + IntToStr(Integer(P5)));
end;

The result is:

P1: 5006500
P2: 1242488
P3: 1242488
P4: 5006500
P5: 67

Where:

- P2 and P3, is the address of Buffer: AnsiString 
- P5 is the Char Ord value of Buffer[1] char, in this case "67 = C"
- How about P1 and P4?

What is the meaning of P1 and P4?

Answer 1

Nice puzzle, but I have the solution:

• P2 and P3 are the address of the pointer to the buffer
• P1 and P4 are the address of the buffer
• P5 is the first element in the buffer

I have added comment in the code:

var
  Buffer: AnsiString;
  P1: Pointer;
  P2: Pointer;
  P3: Pointer;
  P4: Pointer;
  P5: Pointer;
begin
  P1:= PAnsiString(Buffer); 
  (* A cast from AnsiString to PAnsiString has no real meaning 
     because both are a pointer to a block of characters ()
  P2:= Addr(Buffer);
  P3:= @Buffer;
  (* Both Addr and @ give the address of a variable. The variable Buffer is 
     a pointer so we get the address of the pointer, not the value of the 
     pointer. *)
  P4:= Pointer(Buffer);
  (* See the remark on P1. Due to the cast both give the same result. *)
  P5:= PChar(Buffer[1]);
  (* This looks like a pointer to the first element. But the cast changes 
     it into the character. *)
  WriteLn('P1: ' + IntToStr(Integer(P1)));
  WriteLn('P2: ' + IntToStr(Integer(P2)));
  WriteLn('P3: ' + IntToStr(Integer(P3)));
  WriteLn('P4: ' + IntToStr(Integer(P4)));
  WriteLn('P5: ' + IntToStr(Integer(P5)));
end;

Answer 2

An AnsiString is implemented as a pointer. An AnsiString variable holds nothing but an address. The address is that of the first character in the string, or nil if the string is empty.

A PAnsiString is a pointer to an AnsiString variable. It's a pointer to a pointer to the first character of a string. When you say PAnsiString(Buffer), you're telling the compiler to treat the pointer in Buffer as though it were a pointer to an AnsiString instead of a pointer to character data. The address 5006500 is the location of the first character of the string, C.

You have a record in memory that represents the string:

                +-----------+
                | $ffffffff | -1 reference count (4 bytes)
                +-----------+
Buffer:         | $00000001 | length (4 bytes)
+---------+     +-----------+
| 5006500 | --> |       'C' | first character (1 byte)
+---------+     +-----------+
                |        #0 | null terminator (1 byte)
                +-----------+

Buffer holds the address of the byte with C in it. You type-cast that to have type PAnsiString instead of AnsiString. You told the compiler that you had this layout:

                                  +-----------+
                                  |       ... |
                                  +-----------+
Buffer:                           |       ... |
+---------+     +-----------+     +-----------+
| 5006500 | --> | $00000043 | --> |   garbage | first character
+---------+     +-----------+     +-----------+
                                  |       ... |
                                  +-----------+

When I'm reasoning about pointers, I draw diagrams just like this. If you don't keep some paper next to you on your desk, you're doing yourself a disservice.