Little Test Explanation
In the first set circle_area is assigned the object 
, because
before the assignment occurs the variable r on the right side of := is fully evaluated (its pointer is followed), and the found
value, 2, is substituted into the expression. This evaluated expression is then
assigned to circle_area.
The variable r is also fully evaluated in the circle_area
assignment of the second set, but this time r evaluates to r, not 2 (its
pointer points nowhere). Thus, cirlce_area is assigned to the object 
. Every time Maple
encounters an expression containing variables, it evaluates those variables unless told otherwise. As a result, in
the second set, every time the value of r changes, circle_area, assigned the
object 
, evaluates to a
different value. Take a look at the pictures following each set below.
The first set.
> r := 2;
> circle_area := Pi*r^2;
> circle_area;
> r := 8;
> circle_area;
At this point a snippet from memory looks like:
The second set.
> r := 'r';
> circle_area := Pi*r^2;
> r := 2;
> circle_area;
> r := 8;
> circle_area;
And at this point a snippet of memory looks like:
As an aside, note that neither of the expressions pointed to in this picture have pointers associated with them, even though they are both objects. An object is anything that Maple can store in memory, but not all objects can be variables. For example, an integer cannot be a variable, and neither can any expression containing a multiplication operation. Another way to think of an object that can be a variable is that such objects can be placed on the left side of an assignment operator, :=. If you try this with objects that cannot act as variables, Maple will give you an error message. Try it.
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