Chandra, >I have 2 small pieces of code: > >Code - 1 >-------- >type > TDataType = Array[0..3] of char; > >procedure hello; >var > a,b : Array[0..3] of char; > c,d : TDataType; >begin > a := b; > c := d; >end; > > >Code - 2 >-------- >type > TDataType = Array[0..3] of char; > >procedure hello; >var > a : Array[0..3] of char; > b : Array[0..3] of char; > c : TDataType; > d : TDataType; >begin > a := b; > c := d; >end; > >------------- >Code1 gets compiled and runs fine but code2 gives compile time error >in line a := b;
I assume the error is an incompatible type error. Pascal implements "strict typing". This means that the variable types must exactly the same for the compiler to allow assignments. I am guessing that you assume the two separately declared unnamed types Array[0..3] of char; are exactly the same. For scalar variables, this is true. Separately declared integers, for example are always assignment compatible. Structures, on the other hand, like arrays, are not assumed to be identical just because they have the same definition. For example, let's say you have these structures: var EmployeeID: array [0..20] of char; EmployeeAddress: array [0..20] of char; Even though these two types happen to be the same length, a Pascal compiler should enforce strict typing and prevent you from inadvertently assigning an EmployeeID to an EmployeeAddress. Obviously, such an assignment would not make sense. The reason you first example worked for a:=b is because you defined them in the same var declaration, so the compiler could assume they were exactly the same type. The reason c:=d works in both examples is because you used a named type. When a named type is used, the compiler can assume the types are identical. Hope this helps, Glenn Lawler www.incodesystems.com
