Introduction to Parameters and Objects - Building Java Programs: A Back to Basics Approach - page 137

17
18 System.out.print("<");
19 writeSpaces(spaces1 * spaces2 - 5);
20 System.out.println(">");
21 }
22
23 // writes "number" spaces on the current output line
24 public static void writeSpaces( int number) {
25 for ( int i = 1; i <= number; i++) {
26 System.out.print(" ");
27 }
28 }
29 }

In the first two lines of the main method, the computer finds instructions to allocate
and initialize two variables:

spaces1 3 spaces2 5

The next three lines of code produce an output line with three spaces bounded by
asterisks on either side:

System.out.print("*");
writeSpaces(spaces1);
System.out.println("*");

You can see where the asterisks come from, but look at the method call that pro-
duces the spaces. When Java executes the call on writeSpaces , it must set up its
parameter. To set up the parameter, Java first evaluates the expression being passed as
the actual parameter. The expression is simply the variable spaces1 , which has the
value 3 . Therefore, the expression evaluates to 3 . Java uses this result to initialize a
local variable called number .
The following diagram indicates how the computer's memory would look as the
writeSpaces method is entered the first time. Because there are two methods
involved ( main and writeSpaces ), the diagram indicates which variables are local
to main ( spaces1 and spaces2 ) and which are local to writeSpaces (the parameter
number ):

method main method writeSpaces

spaces1 3 spaces2 5 number 3