The with statement serves to access the elements of a record or object or class, without having to specify the name of the each time. The syntax for a with statement is
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With statement
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The variable reference must be a variable of a record, object or class type. In the with statement, any variable reference, or method reference is checked to see if it is a eld or method of the record or object or class. If so, then that eld is accessed, or that method is called. Given the declaration:
Type Passenger = Record
Name : String[30]; Flight : String[10]; end; Var TheCustomer : Passenger; |
The following statements are completely equivalent:
TheCustomer.Name := 'Michael';
TheCustomer.Flight := 'PS901'; |
and
With TheCustomer do
begin Name := 'Michael'; Flight := 'PS901'; end; |
The statement
With A,B,C,D do Statement;
|
is equivalent to
With A do
With B do With C do With D do Statement; |
This also is a clear example of the fact that the variables are tried last to rst, i.e., when the compiler encounters a variable reference, it will rst check if it is a eld or method of the last variable. If not, then it will check the last-but-one, and so on. The following example shows this;
Program testw;
Type AR = record X,Y : Longint; end; PAR = Record; Var S,T : Ar; begin S.X := 1;S.Y := 1; T.X := 2;T.Y := 2; With S,T do WriteLn (X,' ',Y); end. |
The output of this program is
2 2
|
Showing thus that the X,Y in the WriteLn statement match the T record variable.
Remark: When using a With statement with a pointer, or a class, it is not permitted to change the pointer or the class in the With block. With the de nitions of the previous example, the following illustrates what it is about:
Var p : PAR; begin With P^ do begin // Do some operations P:=OtherP; X:=0.0; // Wrong X will be used !! end; |
The reason the pointer cannot be changed is that the address is stored by the compiler in a temporary register. Changing the pointer won't change the temporary address. The same is true for classes.