Announcements Homework P 1 due tomorrow Thursday Homework

Announcements • Homework P 1 due tomorrow (Thursday) • Homework P 2 has been handed out and is on the web CS 100 Lecture 7 1

Today’s Topics • Review • Method calls in (excruciating : -) ) detail. CS 100 Lecture 7 2

Review • Overview of material so far • Scope of variables – fields in classes – parameters of methods – local variables, variables in blocks • The while loop CS 100 Lecture 7 3

Method Calls • Today, we will describe how a method call is executed • You should know this material very well • Many difficulties in programming come from not understanding exactly how statements, method calls, etc. are being executed. CS 100 Lecture 7 4

Executing an Assignment • To execute an assignment x= e; (1) Evaluate e (yielding a value v, say) (2) Store v in the variable described by x. instance of class C b d 7 c __4__ d __3__ • b. c = 2 * b. c value is 8, so store 8 in field c of b • b = new C(8, 7) value is name of new instance of class C, so store that in b CS 100 Lecture 7 5

Method Invocation • 0. Allocate a new set of locations to contain the parameters and local variables of the method being called, in what we call a frame. • 1. Assign the arguments of the call to the parameters of the method. • 2. Execute the procedure body. • 3. Delete the frame produced in step 0; if the method is a function, return the value of the function to the place of call. • Terminology: parameter: formal parameter argument: actual parameter CS 100 Lecture 7 6

Example // Yield the maximum of x and y public int max(int x, int y) {int z; z= x; if (y > x) z= y; return z; } The call is in the assignment given below. int b; b= max (3*20, 6+5) + 2; CS 100 Lecture 7 7

b= max (3*20, b+5) + 2; The state before execution, with b = 7: 7 b The state after execution of step 0. 7 b frame for the call ? x ? y The state after execution of step 1. 7 b x CS 100 60 y 12 Lecture 7 ? z frame for the call z ? 8

Performing step 2, initial state b 60 x After execution of z= x; x 7 60 12 y 7 b y frame for the call ? z frame for the call 12 z 60 After execution of the conditional statement: 7 b frame for the call x CS 100 60 y Lecture 7 12 z 60 9

Executing the return, which terminates execution of the call (perform step 3). b 7 60 b= max (3*20, b+5) + 2; b 62 Memorize the steps in executing a procedure call, given on slide 3. Practice executing simple procedure calls yourself. CS 100 Lecture 7 10

// Add b to c public void add. C(Coordinate b, Coordinate c) {c. x= c. x+b. x; c. y= c. y+b. y; } Execute the call add. C(d, b); assuming the initial state is: Coordinate b x 3 Coordinate d x CS 100 y 5 1 Lecture 7 y 4 11

Step 0: allocate a frame for the method Note that, below, there are two variables named b. There is no ambiguity. Later, when executing the method body, always use the variables in the frame for the call. Coordinate b x 5 y 3 Coordinate d x 1 y 4 frame for call b CS 100 ? Lecture 7 c ? 12

• Execution of step 1 yields: Coordinate b x y 5 3 Coordinate d x 1 y 4 frame for call b CS 100 c Lecture 7 13

• Execution of step 2 yields: Coordinate b x y 6 7 Coordinate d x 1 y 4 frame for call b CS 100 c Lecture 7 14

• Execution of step 3 yields: Coordinate b x 6 7 Coordinate d x CS 100 y 1 Lecture 7 y 4 15

Execution of a new Coordinate & constructor public class Coordinate {public int x; public int y; // Constructor: an instance with x= b and y=0 public Coordinate(int b) {x= b; y= 0; }} Coordinate d; d= new Coordinate(9); To evaluate new Coordinate(9): – 0. Create a new instance of Coordinate – 1. Execute the call Coordinate(9) CS 100 Lecture 7 16

Initial State and Step 0 Initial state: d ? Execution of step 0 yields the following. Note that we have placed method Coordinate within the instance. The instance contains all the fields and methods defined within the class. This will be important when executing the method body, as shown below, to follow the scope rules. Coordinate x ? y ? Coordinate (constructor) CS 100 Lecture 7 17

Step 1: Execute the call on the constructor. Rule: when drawing the frame, place the frame within the class instance that contains the called method! Coordinate x y ? ? Coordinate (constructor) frame for call b Coordinate ? x y ? ? Coordinate (constructor) frame for call b CS 100 Lecture 7 9 18

After execution of constructor body Coordinate y 9 x 0 Coordinate (constructor) frame for call b 9 After call is completed: d= new Coordinate(9); Coordinate x 9 y 0 Coordinate (constructor) So the name of this instance is stored in d. CS 100 Lecture 7 19

Call by Value • All of this is to say that all parameters to Java methods are “call by value” • If you pass a boolean, e. g. to a method, it’s parameter is a copy of whatever value was being passed • The method does not change the original variable’s value CS 100 Lecture 7 20

Example -- call by value Class pass. By. Value { public static void main(String[] args) { double one = 1. 0 System. out. println(“before: one = ” + one); halve. It(one); System. out. println(“after: one = ” + one); } public static void halve. It(double arg) { arg = arg / 2. 0; // divide arg by two System. out. println(“halved: arg = ” + arg); } CS 100 Lecture 7 } 21

Slightly different for objects class pass. Ref. By. Value { public static void main(String[] args) { Body sirius = new Body(“Sirius, null); System. out. println(“before: ” + sirius); common. Name(sirius): System. out. println(“after: ” + sirius); } public static void common. Name(Body body. Ref) { body. Ref. name = “Dog Star”; body. Ref = null; } } CS 100 Lecture 7 22

Output of previous before: 0 (Sirius) after: 0 (Dog Star) • The contents of the object have been modified, but. . . • . . . the reference body. Ref still refers to the Body object, even though common. Name changed its value to null • Remember: aliases of objects CS 100 Lecture 7 23

this -- briefly • Commonly used to pass a reference to the current object to other methods • The this reference always refers the current object that is executing the code • Suppose you want to add the current object to a list of objects w/in a method called on that object: Service. add(this); CS 100 Lecture 7 24

Increment and Decrement • • ++ and -a++ is equivalent to a = a + 1 ++a is equivalent to a = a + 1 Difference is when the value is returned: if prefix, the operation is applied before the value is return, if postfix, after • Example a = 16; System. out. println(++a + “ ” + a++ + “ ” + a); CS 100 Lecture 7 25

Assignment Operators • Any arithmetic or binary operator can be concatenated with = to form another assignment operator • a *= b + 1; // same as a = a * (b + 1) • a += 5; // same as a = a + 5; • var op= expr; is equivalent to var = ((var) op (expr)); CS 100 Lecture 7 26