Stefan Andrei
description
Transcript of Stefan Andrei
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Programming Fundamentals I (COSC-1336), Lecture 5 (prepared after Chapter 5 of Liang’s 2011 textbook)Stefan Andrei
04/21/23 1COSC-1336, Lecture 5
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Overview of Previous Lecture To write programs for executing statements repeatedly
using a while loop (§4.2). To develop a program for GuessNumber and
SubtractionQuizLoop (§4.2.1). To follow the loop design strategy to develop loops
(§4.2.2). To develop a program for SubtractionQuizLoop
(§4.2.3). To control a loop with a sentinel value (§4.2.3). To obtain large input from a file using input redirection
rather than typing from the keyboard (§4.2.4). To write loops using do-while statements (§4.3).
04/21/23 2COSC-1336, Lecture 5
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Overview of Previous Lecture (cont) To write loops using for statements (§4.4). To discover the similarities and differences of three
types of loop statements (§4.5). To write nested loops (§4.6). To learn the techniques for minimizing numerical errors
(§4.7). To learn loops from a variety of examples (GCD,
FutureTuition, MonteCarloSimulation) (§4.8). To implement program control with break and continue
(§4.9). (GUI) To control a loop with a confirmation dialog
(§4.10).
04/21/23 3COSC-1336, Lecture 5
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Motivation of the current lecture Find the sum of integers from 1 to 10, from 20 to 30, and from 35 to 45, respectively.
04/21/23 4COSC-1336, Lecture 5
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First Attempt
int sum = 0;for (int i = 1; i <= 10; i++) sum += i;System.out.println("Sum from 1 to 10 is " + sum);
sum = 0;for (int i = 20; i <= 30; i++) sum += i;System.out.println("Sum from 20 to 30 is " + sum);
sum = 0;for (int i = 35; i <= 45; i++) sum += i;System.out.println("Sum from 35 to 45 is " + sum);
04/21/23 5COSC-1336, Lecture 5
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Problemint sum = 0;for (int i = 1; i <= 10; i++) sum += i;System.out.println("Sum from 1 to 10 is " + sum);
sum = 0;for (int i = 20; i <= 30; i++) sum += i;System.out.println("Sum from 20 to 30 is " + sum);
sum = 0;for (int i = 35; i <= 45; i++) sum += i;System.out.println("Sum from 35 to 45 is " + sum);
04/21/23 6COSC-1336, Lecture 5
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Solutionpublic static int sum(int i1, int i2) { int sum = 0; for (int i = i1; i <= i2; i++) sum += i; return sum;}
public static void main(String[] args) { System.out.println("Sum from 1 to 10 is " + sum(1, 10)); System.out.println("Sum from 20 to 30 is " + sum(20, 30)); System.out.println("Sum from 35 to 45 is " + sum(35, 45));}
04/21/23 7COSC-1336, Lecture 5
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Overview of This Lecture To define methods, invoke methods, and pass
arguments to a method (§5.2-5.5). To develop reusable code that is modular, easy-to-
read, easy-to-debug, and easy-to-maintain. (§5.6). To use method overloading and understand
ambiguous overloading (§5.7). To design and implement overloaded methods
(§5.8).
04/21/23 8COSC-1336, Lecture 5
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Overview of This Lecture (cont.) To determine the scope of variables (§5.9). To know how to use the methods in the Math
class (§§5.10-5.11). To learn the concept of method abstraction
(§5.12). To design and implement methods using
stepwise refinement (§5.12).
04/21/23 9COSC-1336, Lecture 5
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Defining MethodsA method is a collection of statements that are grouped together to perform an operation.
public static int max(int num1, int num2) {
int result; if (num1 > num2) result = num1; else result = num2; return result;
}
modifier return value
type method name
formal parameters
return value
method body
method header
parameter list
Define a method Invoke a method
int z = max(x, y);
actual parameters (arguments)
method signature
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Method SignatureMethod signature is the combination of the method name and the parameter list.
public static int max(int num1, int num2) {
int result; if (num1 > num2) result = num1; else result = num2; return result;
}
modifier return value
type method name
formal parameters
return value
method body
method header
parameter list
Define a method Invoke a method
int z = max(x, y);
actual parameters (arguments)
method signature
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Formal ParametersThe variables defined in the method header are known as formal parameters.
public static int max(int num1, int num2) {
int result; if (num1 > num2) result = num1; else result = num2; return result;
}
modifier return value
type method name
formal parameters
return value
method body
method header
parameter list
Define a method Invoke a method
int z = max(x, y);
actual parameters (arguments)
method signature
04/21/23 12COSC-1336, Lecture 5
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Actual Parameters When a method is invoked, you pass a value to the parameter.
This value is referred to as actual parameter or argument.
public static int max(int num1, int num2) {
int result; if (num1 > num2) result = num1; else result = num2; return result;
}
modifier return value
type method name
formal parameters
return value
method body
method header
parameter list
Define a method Invoke a method
int z = max(x, y);
actual parameters (arguments)
method signature
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Return Value Type A method may return a value.
The return value type is the data type of the value the method returns.
If the method does not return a value, the return value type is the keyword void.
For example, the return value type in the main() method is void.
public static int max(int num1, int num2) {
int result; if (num1 > num2) result = num1; else result = num2; return result;
}
modifier return value
type method name
formal parameters
return value
method body
method header
parameter list
Define a method Invoke a method
int z = max(x, y);
actual parameters (arguments)
method signature
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Calling Methods
Testing the max() method;
This program demonstrates calling a method max to return the largest of the int values.
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Calling Methods (cont.)
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
pass the value of i pass the value of j
animation
04/21/23 16COSC-1336, Lecture 5
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Trace Method Invocation
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
i is now 5
animation
04/21/23 17COSC-1336, Lecture 5
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Trace Method Invocation
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
j is now 2
animation
04/21/23 18COSC-1336, Lecture 5
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Trace Method Invocation
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
invoke max(i, j)
animation
04/21/23 19COSC-1336, Lecture 5
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Trace Method Invocation
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
invoke max(i, j)Pass the value of i to num1Pass the value of j to num2
animation
04/21/23 20COSC-1336, Lecture 5
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Trace Method Invocation
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
declare variable result
animation
04/21/23 21COSC-1336, Lecture 5
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Trace Method Invocation
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
(num1 > num2) is true since num1 is 5 and num2 is 2
animation
04/21/23 22COSC-1336, Lecture 5
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Trace Method Invocation
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
result is now 5
animation
04/21/23 23COSC-1336, Lecture 5
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Trace Method Invocation
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
return result, which is 5
animation
04/21/23 24COSC-1336, Lecture 5
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Trace Method Invocation
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
return max(i, j) and assign the return value to k
animation
04/21/23 25COSC-1336, Lecture 5
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Trace Method Invocation
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
Call the println() method
animation
04/21/23 26COSC-1336, Lecture 5
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CAUTION A return statement is required for a value-returning method. The method shown below in (a) is logically correct, but it has a compilation error because the Java compiler thinks it possible that this method does not return any value.
To fix this problem, delete if (n < 0) in (a), so that the compiler will see a return statement to be reached regardless of how the if statement is evaluated.
public static int sign(int n) { if (n > 0) return 1; else if (n == 0) return 0; else if (n < 0) return –1; }
(a)
Should be
(b)
public static int sign(int n) { if (n > 0) return 1; else if (n == 0) return 0; else return –1; }
04/21/23 27COSC-1336, Lecture 5
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Reuse Methods from Other Classes NOTE: One of the benefits of methods is for reuse. The max() method can be invoked from any class besides TestMax. If you create a new class Test, you can invoke the max() method using ClassName.methodName (e.g., TestMax.max).
04/21/23 28COSC-1336, Lecture 5
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Call Stacks
(a) The main method is invoked.
Space required for the main method k: j: 2 i: 5
(b) The max method is invoked.
Space required for the max method num2: 2 num1: 5
(d) The max method is finished and the return value is sent to k.
(e) The main method is finished.
Stack is empty
Space required for the main method k: j: 2 i: 5
Space required for the main method k: 5 j: 2 i: 5
(c) The max method is being executed.
Space required for the max method result: 5 num2: 2 num1: 5
Space required for the main method k: j: 2 i: 5
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Trace Call Stack
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
i is declared and initialized
The main method is invoked.
i: 5
animation
04/21/23 30COSC-1336, Lecture 5
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Trace Call Stack
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
j is declared and initialized
The main method is invoked.
j: 2 i: 5
animation
04/21/23 31COSC-1336, Lecture 5
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Trace Call Stack
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
Declare k
The main method is invoked.
Space required for the main method
k: j: 2 i: 5
animation
04/21/23 32COSC-1336, Lecture 5
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Trace Call Stack
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
Invoke max(i, j)
The main method is invoked.
Space required for the main method
k: j: 2 i: 5
animation
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Trace Call Stack
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
pass the values of i and j to num1 and num2
The max method is invoked.
num2: 2 num1: 5
Space required for the main method
k: j: 2 i: 5
animation
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Trace Call Stack
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
declare variable result
The max method is invoked.
result:
num2: 2 num1: 5
Space required for the main method
k: j: 2 i: 5
animation
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Trace Call Stack
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
(num1 > num2) is true
The max method is invoked.
result:
num2: 2 num1: 5
Space required for the main method
k: j: 2 i: 5
animation
04/21/23 36COSC-1336, Lecture 5
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Trace Call Stack
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
Assign num1 to result
The max method is invoked.
Space required for the max method result: 5
num2: 2 num1: 5
Space required for the main method
k: j: 2 i: 5
animation
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Trace Call Stack
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
Return result and assign it to k
The max method is invoked.
Space required for the max method result: 5
num2: 2 num1: 5
Space required for the main method k:5
j: 2 i: 5
animation
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Trace Call Stack
public static void main(String[] args) { int i = 5; int j = 2; int k = max(i, j); System.out.println( "The maximum between " + i + " and " + j + " is " + k); }
public static int max(int num1, int num2) { int result; if (num1 > num2) result = num1; else result = num2; return result; }
Call the println() method
The main method is invoked.
Space required for the main method k:5
j: 2 i: 5
animation
04/21/23 39COSC-1336, Lecture 5
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void Method Example
This type of method does not return a value. The method performs some actions.
public class TestVoidMethod { public static void main(String[] args) { System.out.print("The grade is "); printGrade(78.5); System.out.print("The grade is "); printGrade(59.5); } // . . . public static void printGrade() }
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Method public static void printGrade()public static void printGrade(double score) {
if (score >= 90.0) {
System.out.println('A'); }
else if (score >= 80.0) {
System.out.println('B'); }
else if (score >= 70.0) {
System.out.println('C'); }
else if (score >= 60.0) {
System.out.println('D'); }
else { System.out.println('F'); }
} 04/21/23 41COSC-1336, Lecture 5
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Compiling and running TestVoidMethod.java
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Passing Parameterspublic static void nPrintln(String message, int n) { for (int i = 0; i < n; i++) System.out.println(message);}
Suppose you invoke the method using nPrintln("Welcome to Java", 5); What is the output? Suppose you invoke the method using nPrintln("Computer Science", 15); What is the output?
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Pass by Value This program demonstrates passing values to the methods.public class Increment { public static void main(String[] args) { int x = 1; System.out.println("Before the call, x is " + x); increment(x); System.out.println("after the call, x is " + x); } public static void increment(int n) { n++; System.out.println("n inside the method is " + n); } }
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Compiling and running Increment.java
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Pass by Value Testing Pass by value.
This program demonstrates passing values to the methods.public class TestPassByValue { public static void main(String[] args) { int num1 = 1, num2 = 2; System.out.println("Before invoking the swap method, num1 is " + num1 + " and num2 is " + num2); swap(num1, num2); System.out.println("After invoking the swap method, num1 is " + num1 + " and num2 is " + num2); } // . . . }
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public static void swap()
/** Swap two variables */
public static void swap(int n1, int n2) {
System.out.println("\tInside the swap method");
System.out.println("\t\tBefore swapping n1 is "
+ n1 + " n2 is " + n2);
int temp = n1;
n1 = n2;
n2 = temp;
System.out.println("\t\tAfter swapping n1 is "
+ n1 + " n2 is " + n2);
}
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Compiling and running TestPassByValue.java
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Pass by Value (cont.)
The main method is invoked
The values of num1 and num2 are passed to n1 and n2. Executing swap does not affect num1 and num2.
Space required for the main method
num2: 2 num1: 1
The swap method is invoked
Space required for the main method
num2: 2 num1: 1
Space required for the swap method temp:
n2: 2 n1: 1
The swap method is finished
Space required for the main method
num2: 2 num1: 1
The main method is finished
Stack is empty
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Modularizing Code Methods can be used to reduce redundant coding and enable code reuse. Methods can also be used to modularize code and improve the quality of the program.
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Method Overloading Method overloading is the process of giving a single
method name multiple definitions. If a method is overloaded, the method name is not
sufficient to determine which method is being called. The signature of each overloaded method must be
unique. The signature includes the number, type, and order
of the parameters. The return type of the method is not part of the
signature. That is, overloaded methods cannot differ only by
their return type.
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Method Overloading The compiler determines which method is being
invoked by analyzing the parameters (the first signature is (int) and the second one is (int, double):
float method1(int x){ return x + .375;}
float method1(int x, double y){ return x * y;}
result = method1(25, 4.32)
Invocation
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Method Overloading
The println() method is overloaded: println(String s) println(int i) println(double d)
and so on...
The following lines invoke different versions of the println() method:
System.out.println("The total is:"); System.out.println(total);
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Overloading Methods
Overloading the max() method:
public static double max(double num1, double num2) { if (num1 > num2) return num1; else return num2;}
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TestMethodOverloading.java
public class TestMethodOverloading {
public static void main(String[] args) {
System.out.println("The maximum between 3 and 4 is " + max(3, 4));
System.out.println("The maximum between 3.0 and 5.4 is " + max(3.0, 5.4));
System.out.println("The maximum between 3.0, 5.4, and 10.14 is " + max(3.0, 5.4, 10.14));
}
// . . . The max() methods come next
}
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TestMethodOverloading.java (cont.)
public static int max(int num1, int num2) { if (num1 > num2) return num1; else return num2; } public static double max(double num1, double num2){ if (num1 > num2) return num1; else return num2; } public static double max(double num1, double num2,
double num3) { return max(max(num1, num2), num3); }
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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.
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Ambiguous Invocationpublic 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; }}
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Type error during compilation
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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 can be used.
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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.
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Scope of Local Variables (cont.) A variable declared in the initial action part of a for loop header has its scope in the entire loop. But a variable declared inside a for loop body has its scope limited in the loop body from its declaration and to the end of the block that contains the variable.
public static void method1() { . . for (int i = 1; i < 10; i++) { . . int j; . . . } }
The scope of j
The scope of i
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Scope of Local Variables (cont.)
public static void method1() { int x = 1; int y = 1;
for (int i = 1; i < 10; i++) {
x += i; }
for (int i = 1; i < 10; i++) {
y += i; } }
It is fine to declare i in two non-nesting blocks
public static void method2() { int i = 1; int sum = 0;
for (int i = 1; i < 10; i++) {
sum += i; } }
It is wrong to declare i in two nesting blocks
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Scope of Local Variables (cont.)// Fine with no errorspublic 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; }}
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Scope of Local Variables (cont.)// With no errorspublic static void incorrectMethod() { int x = 1; int y = 1; for (int i = 1; i < 10; i++) { int x = 0; x += i; }}
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Method Abstraction You can think of the method body as a black box that contains the detailed implementation for the method.
Method Header
Method body
Black Box
Optional arguments for Input
Optional return value
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Benefits of Methods
Write a method once and reuse it anywhere.Information hiding: hide the implementation from the user.Reduce complexity.
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The Math Class
Class constants: PI E
Class methods: Trigonometric Methods Exponent Methods Rounding Methods min(), max(), abs(), and random() Methods
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Trigonometric Methods
sin(double a)
cos(double a)
tan(double a)
asin(double a)
acos(double a)
atan(double a)
Radians
toRadians(90)
Examples:Math.sin(0) returns 0.0
Math.sin(Math.PI / 6) returns 0.5
Math.sin(Math.PI / 2) returns 1.0
Math.cos(0) returns 1.0
Math.cos(Math.PI / 6) returns 0.866
Math.cos(Math.PI / 2) returns 0
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Exponent Methods exp(double a)
Returns e raised to the power of a.
log(double a)
Returns the natural logarithm of a.
log10(double a)
Returns the 10-based 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.
Examples:
Math.exp(1) returns 2.71
Math.log(2.71) returns 1.0
Math.pow(2, 3) returns 8.0
Math.pow(3, 2) returns 9.0
Math.pow(3.5, 2.5) returns 22.91765
Math.sqrt(4) returns 2.0
Math.sqrt(10.5) returns 3.24
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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).
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Rounding Methods ExamplesMath.ceil(2.1) returns 3.0
Math.ceil(2.0) returns 2.0Math.ceil(-2.0) returns –2.0Math.ceil(-2.1) returns -2.0Math.floor(2.1) returns 2.0Math.floor(2.0) returns 2.0Math.floor(-2.0) returns –2.0Math.floor(-2.1) returns -3.0Math.rint(2.1) returns 2.0
Math.rint(2.0) returns 2.0Math.rint(-2.0) returns –2.0Math.rint(-2.1) returns -2.0Math.rint(2.5) returns 2.0Math.rint(-2.5) returns -2.0Math.round(2.6f) returns 3 Math.round(2.0) returns 2 Math.round(-2.0f) returns -2 Math.round(-2.6) returns -3
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min(), max(), and abs()
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).
Examples:
Math.max(2, 3) returns 3
Math.max(2.5, 3) returns 3.0
Math.min(2.5, 3.6) returns 2.5
Math.abs(-2) returns 2
Math.abs(-2.1) returns 2.1
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The random() Method Generates a random double value greater than or equal to 0.0 and less than 1.0 (0 <= Math.random() < 1.0).
Examples:
(int)(Math.random() * 10) Returns a random integer between 0 and 9.
50 + (int)(Math.random() * 50) Returns a random integer between 50 and 99.
In general,
a + Math.random() * b Returns a random number between
a and a + b, excluding a + b.
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Case Study: Generating Random Characters Computer programs process numerical data and characters. You have seen many examples that involve numerical data. It is also important to understand characters and how to process them. As introduced in Section 2.9, each character has a unique Unicode between 0 and FFFF in hexadecimal (65535 in decimal).
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Case Study: Generating Random Characters (cont.) To generate a random character is to generate a random integer between 0 and 65535 using the following expression:
note that since 0 <= Math.random() < 1.0, you have to add 1 to 65535. (int)(Math.random() * (65535 + 1))
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Case Study: Generating Random Characters (cont.) Now let us consider how to generate a random lowercase letter. The Unicode for lowercase letters are consecutive integers starting from the Unicode for 'a', then for 'b', 'c', ..., and 'z'. The Unicode for 'a' is (int)'a'. So, a random integer between (int)'a' and (int)'z' is:
(int)((int)'a' + Math.random() *
((int)'z' – (int)'a' + 1)
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Case Study: Generating Random Characters (cont.) As discussed in Chapter 2, all numeric operators can be applied to the char operands. The char operand is cast into a number if the other operand is a number or a character. So, the preceding expression can be simplified as follows: 'a' + Math.random() * ('z' - 'a' + 1)
So a random lowercase letter is(char)('a' + Math.random() * ('z' - 'a' + 1))
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Case Study: Generating Random Characters (cont.) To generalize the foregoing discussion, a random character between any two characters ch1 and ch2 with ch1 < ch2 can be generated as follows:
(char)(ch1 + Math.random() * (ch2 – ch1 + 1))
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The RandomCharacter Class// RandomCharacter.java: Generate random characterspublic class RandomCharacter { /** Generate a random character between ch1 and ch2 */ public static char getRandomCharacter(char ch1, char ch2) { return (char)(ch1 + Math.random() * (ch2 - ch1 + 1)); } /** Generate a random lowercase letter */ public static char getRandomLowerCaseLetter() { return getRandomCharacter('a', 'z'); } /** Generate a random uppercase letter */ public static char getRandomUpperCaseLetter() { return getRandomCharacter('A', 'Z'); } /** Generate a random digit character */ public static char getRandomDigitCharacter() { return getRandomCharacter('0', '9'); } /** Generate a random character */ public static char getRandomCharacter() { return getRandomCharacter('\u0000', '\uFFFF'); }}
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Stepwise Refinement (Optional) The concept of method abstraction can be applied to the process of developing programs. When writing a large program, you can use the “divide and conquer” strategy, also known as stepwise refinement, to decompose it into subproblems. The subproblems can be further decomposed into smaller, more manageable problems.
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PrintCalendar Case Study Let us use the PrintCalendar example to demonstrate the stepwise refinement approach.
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Design Diagram
printCalendar (main)
readInput printMonth
getStartDay
printMonthTitle printMonthBody
getTotalNumOfDays
getNumOfDaysInMonth
getMonthName
isLeapYear
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Design Diagram
printCalendar (main)
readInput printMonth
getStartDay
printMonthTitle printMonthBody
getTotalNumOfDays
getNumOfDaysInMonth
getMonthName
isLeapYear
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Design Diagram
printCalendar (main)
readInput printMonth
getStartDay
printMonthTitle printMonthBody
getTotalNumOfDays
getNumOfDaysInMonth
getMonthName
isLeapYear
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Design Diagram
printCalendar (main)
readInput printMonth
getStartDay
printMonthTitle printMonthBody
getTotalNumOfDays
getNumOfDaysInMonth
getMonthName
isLeapYear
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Design Diagram
printCalendar (main)
readInput printMonth
getStartDay
printMonthTitle printMonthBody
getTotalNumOfDays
getNumOfDaysInMonth
getMonthName
isLeapYear
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Design Diagram
printCalendar (main)
readInput printMonth
getStartDay
printMonthTitle printMonthBody
getTotalNumOfDays
getNumOfDaysInMonth
getMonthName
isLeapYear
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Implementation: Top-Down Top-down approach is to implement one method in the structure chart at a time from the top to the bottom. Stubs can be used for the methods waiting to be implemented. A stub is a simple but incomplete version of a method. The use of stubs enables you to test invoking the method from a caller. Implement the main method first and then use a stub for the printMonth() method. For example, let printMonth() display the year and the month in the stub.
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Implementation: Bottom-Up
Bottom-up approach is to implement one method in the structure chart at a time from the bottom to the top. For each method implemented, write a test program to test it. Both top-down and bottom-up methods are fine. Both approaches implement the methods incrementally and help to isolate programming errors and makes debugging easy. Sometimes, they can be used together.
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Summary To define methods, invoke methods, and pass
arguments to a method (§5.2-5.5). To develop reusable code that is modular, easy-to-
read, easy-to-debug, and easy-to-maintain. (§5.6). To use method overloading and understand
ambiguous overloading (§5.7). To design and implement overloaded methods
(§5.8).
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Summary (cont.) To determine the scope of variables (§5.9). To know how to use the methods in the Math
class (§§5.10-5.11). To learn the concept of method abstraction
(§5.12). To design and implement methods using
stepwise refinement (§5.12).
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Reading suggestions
From [Liang: Introduction to Java programming: Eight Edition, 2011 Pearson Education, 0132130807]
Chapter 5 (Methods)
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Coming up next
From [Liang: Introduction to Java programming: Eight Edition, 2011 Pearson Education, 0132130807] Chapter 9 (Strings)
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Thank you for your attention!
Questions?
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