With the list assignment (see List Assignment, Add, Append) it is possible to change a list. The ability to change an object is only available for lists and records. This section describes the semantic consequences of this fact.
You may think that in the following example the second assignment changes the integer, and that therefore the above sentence, which claimed that only lists and records can be changed is wrong
i := 3; i := i + 1;
But in this example not the integer
3 is changed by adding one to it.
Instead the variable
i is changed by assigning the value of
which happens to be
i. The same thing happens in the following
l := [ 1, 2 ]; l := [ 1, 2, 3 ];
The second assignment does not change the first list, instead it assigns
a new list to the variable
l. On the other hand, in the following
example the list is changed by the second assignment.
l := [ 1, 2 ]; l := 3;
To understand the difference first think of a variable as a name for an
object. The important point is that a list can have several names at the
same time. An assignment
var := list; means in this
interpretation that var is a name for the object list. At the end of
the following example
l2 still has the value
[ 1, 2 ] as this list
has not been changed and nothing else has been assigned to it.
l1 := [ 1, 2 ]; l2 := l1; l1 := [ 1, 2, 3 ];
But after the following example the list for which
l2 is a name has
been changed and thus the value of
l2 is now
[ 1, 2, 3 ].
l1 := [ 1, 2 ]; l2 := l1; l1 := 3;
We shall say that two lists are identical if changing one of them by a list assignment also changes the other one. This is slightly incorrect, because if two lists are identical, there are actually only two names for one list. However, the correct usage would be very awkward and would only add to the confusion. Note that two identical lists must be equal, because there is only one list with two different names. Thus identity is an equivalence relation that is a refinement of equality.
Let us now consider under which circumstances two lists are identical.
If you enter a list literal than the list denoted by this literal is a
new list that is not identical to any other list. Thus in the following
l2 are not identical, though they are equal of course.
l1 := [ 1, 2 ]; l2 := [ 1, 2 ];
Also in the following example, no lists in the list
l are identical.
l := ; for i in [1..10] do l[i] := [ 1, 2 ]; od;
If you assign a list to a variable no new list is created. Thus the list
value of the variable on the left hand side and the list on the right
hand side of the assignment are identical. So in the following example
l2 are identical lists.
l1 := [ 1, 2 ]; l2 := l1;
If you pass a list as argument, the old list and the argument of the
function are identical. Also if you return a list from a function, the
old list and the value of the function call are identical. So in the
l2 are identical list
l1 := [ 1, 2 ]; f := function ( l ) return l; end; l2 := f( l1 );
ShallowCopy (see Copy and ShallowCopy)
accept a list and return a new list that is equal to the old list but
that is not identical to the old list. The difference between
ShallowCopy is that in the case of
ShallowCopy the corresponding
elements of the new and the old lists will be identical, whereas in the
Copy they will only be equal. So in the following example
l2 are not identical lists.
l1 := [ 1, 2 ]; l2 := Copy( l1 );
If you change a list it keeps its identity. Thus if two lists are
identical and you change one of them, you also change the other, and they
are still identical afterwards. On the other hand, two lists that are
not identical will never become identical if you change one of them. So
in the following example both
l2 are changed, and are still
l1 := [ 1, 2 ]; l2 := l1; l1 := 2;
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