#### Transcript Chapter 4 Methods

Chapter 4 Methods Introducing Methods – Benefits of methods, Declaring Methods, and Calling Methods Passing Parameters – Pass by Value Overloading Methods – Ambiguous Invocation Scope of Local Variables Method Abstraction The Math Class Case Studies Recursion (Optional) Introducing Methods A method is a collection of statements that are grouped together to perform an operation. Method Structure Introducing Methods, cont. •parameter profile refers to the type, order, and number of the parameters of a method. •method signature is the combination of the method name and the parameter profiles. •The parameters defined in the method header are known as formal parameters. •When a method is invoked, its Declaring Methods public static int max(int num1, int num2) { if (num1 > num2) return num1; else return num2; } Calling Methods Example 4.1 Testing the max method This program demonstrates calling a method max to return the largest of the int values TestMax Run Calling Methods, cont. pass i pass j public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; System.out.println( "The maximum between " + i + " and " + j + " is " + k); } return result; } Calling Methods, cont. The main method i: 5 pass 5 pass 2 The max method num1: 5 parameters j: 2 num2: 2 k: 5 result: 5 CAUTION A return statement is required for a nonvoid method. The following method is logically correct, but it has a compilation error, because the Java compiler thinks it possible that this method does not return any value. public static int xMethod(int n) { if (n > 0) return 1; else if (n == 0) return 0; else if (n < 0) return –1; } To fix this problem, delete if (n<0) in the code. Passing Parameters public static void nPrintln(String message, int n) { for (int i = 0; i < n; i++) System.out.println(message); } Pass by Value Example 4.2 Testing Pass by value This program demonstrates passing values to the methods. TestPassByValue Run Pass by Value, cont. Invoke swap swap(num1, num2) num1 1 num2 2 n1 1 n2 2 The values of num1 and num2 are passed to n1 and n2. Executing swap does not affect num1 and num2. Pass by value swap( n1, n2) Execute swap inside the swap body Swap n1 2 n2 1 temp 1 Overloading Methods Example 4.3 Overloading the max Method public static double max(double num1, double num2) { if (num1 > num2) return num1; else return num2; } TestMethodOverloading Run Ambiguous Invocation Sometimes there may be two or more possible matches for an invocation of a method, but the compiler cannot determine the most specific match. This is referred to as ambiguous invocation. Ambiguous invocation is a compilation error. Ambiguous Invocation public class AmbiguousOverloading { public static void main(String[] args) { System.out.println(max(1, 2)); } public static double max(int num1, double num2) { if (num1 > num2) return num1; else return num2; } public static double max(double num1, int num2) { if (num1 > num2) return num1; else return num2; } Scope of Local Variables A local variable: a variable defined inside a method. Scope: the part of the program where the variable can be referenced. The scope of a local variable starts from its declaration and continues to the end of the block that contains the variable. A local variable must be declared before it Scope of Local Variables, cont. You can declare a local variable with the same name multiple times in different non-nesting blocks in a method, but you cannot declare a local variable twice in nested blocks. Thus, the following code is correct. Scope of Local Variables, cont. // Fine with no errors public static void correctMethod() { int x = 1; int y = 1; // i is declared for (int i = 1; i < 10; i++) { x += i; } // i is declared again for (int i = 1; i < 10; i++) { y += i; } Scope of Local Variables, cont. // With no errors public static void incorrectMethod() { int x = 1; int y = 1; for (int i = 1; i < 10; i++) { int x = 0; x += i; } } Method Abstraction You can think of the method body as a black box that contains the detailed implementation for the method. Optional return value Optional Input Method Signature Method body Black Box Benefits of Methods • Write once and reuse it any times. • Information hiding. Hide the implementation from the user. • Reduce complexity. The Math Class Class constants: – PI –E Class methods: – Trigonometric Methods – Exponent Methods – Rounding Methods – min, max, abs, and random Methods Trigonometric Methods sin(double a) cos(double a) tan(double a) acos(double a) asin(double a) atan(double a) Exponent Methods exp(double a) Returns e raised to the power of a. log(double a) Returns the natural logarithm of a. pow(double a, double b) Returns a raised to the power of b. sqrt(double a) Returns the square root of a. Rounding Methods double ceil(double x) x rounded up to its nearest integer. This integer is returned as a double value. double floor(double x) x is rounded down to its nearest integer. This integer is returned as a double value. double rint(double x) x is rounded to its nearest integer. If x is equally close to two integers, the even one is returned as a double. int round(float x) Return (int)Math.floor(x+0.5). long round(double x) Return (long)Math.floor(x+0.5). min, max, abs, and random max(a, b)and min(a, b) Returns the maximum or minimum of two parameters. abs(a) Returns the absolute value of the parameter. random() Returns a random double value in the range [0.0, 1.0). Example 4.4 Computing Mean and Standard Deviation Generate 10 random numbers and compute the mean and standard deviation n m ean xi n n i 1 n deviation ComputeMeanDeviation x i 1 2 i ( xi ) 2 i 1 n n 1 Run Example 4.5 Obtaining Random Characters Write the methods for generating random characters. The program uses these methods to generate 175 random characters between ‘!' and ‘~' and displays 25 characters per line. To find out the characters between ‘!' and ‘~', see Appendix B, “The ASCII Character Set.” RandomCharacter Run Example 4.5 Obtaining Random Characters, cont. Appendix B: ASCII Character Set Case Studies Example 4.6 Displaying Calendars The program reads in the month and year and displays the calendar for a given month of the year. PrintCalendar Run Design Diagram printCalendar (main) printMonth readInput getStartDay printMonthTitle getTotalNumOfDays getMonthName getNumOfDaysInMonth isLeapYear printMonthBody Recursion (Optional) Example 4.7 Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); ComputeFactorial Run Example 4.7 Computing Factorial, cont. main method: factorial(4) Step 9: factorial(4) returns 24 (4*6) factorial(4) is called in the main factorial(4) = 4*factorial(3) Step 8: factorial(3) returns 6 (3*2) Step 1: factorial(4) calls factorial(3) factorial(3) = 3*factorial(2) Step 7: factorial(2) returns 2 (2*1) Step 2: factorial(3) calls factorial(2) factorial(2) = 2*factorial(1) Step 6: factorial(1) returns 1 (1*1) Step 3: factorial(2) calls factorial(1) factorial(1) = 1*factorial(0) Step 5: factorial(0) returns 1 Step 4: factorial(1) calls factorial(0) factorial(0) = 1 Example 4.7 Computing Factorial, cont. Required 5 Space for factorial(0) Required 1 Space for factorial(4) Required 4 Space for factorial(1) Space Required for factorial(1) Required 3 Space for factorial(2) Space Required for factorial(2) Space Required for factorial(2) Required 2 Space for factorial(3) Space Required for factorial(3) Space Required for factorial(3) Space Required for factorial(3) Space Required for factorial(4) Space Required for factorial(4) Space Required for factorial(4) Space Required for factorial(4) Required 6 Space for factorial(1) Space Required for factorial(2) Required 7 Space for factorial(2) Space Required for factorial(3) Space Required for factorial(3) Required 8 Space for factorial(3) Space Required for factorial(4) Space Required for factorial(4) Space Required for factorial(4) Required 9 Space for factorial(4) Fibonacci Numbers Example 4.8 Computing Finonacci Numbers 0 1 1 2 3 5 8 13 21 34 55 89… f0 f1 fib(2) = fib(0) + fib(1); fib(0) = 0; fib(1) = 1; fib(n) = fib(n-2) + fib(n-1); n>=2 Fibonacci Numbers, cont ComputeFibonacci Run Fibonnaci Numbers, cont. 1 fib(4)= fib(3) + fib(2) call fib(3) return fib(3) 2 call fib(2) fib(3)= fib(2) + fib(1) 7 return fib(2) 3 fib(2)= fib(1) + fib(0) return fib(1) 4 fib(1)= 1 return fib(1) 6 call fib(1) 5 fib(0)= 0 fib(2)= fib(1) + fib(0) fib(1)= 1 8 fib(1)= 1 9 fib(0)= 1 Towers of Hanoi Example 4.9 Solving the Towers of Hanoi Problem Solve the towers of Hanoi problem. TowersOfHanoi Run Towers of Hanoi, cont. A B C A Step 0: Starting status A B B C A B Step 3: Move disk 1 from B to C B C Step 5: Move disk 1 from C to A C A Step 2: Move disk 2 from A to C A C Step 4: Move disk 3 from A to B Step 1: Move disk 1 from A to B A B B C Step 6: Move disk 2 from C to B C A B Step 7: Mve disk 1 from A to B C