cs1101 group1

CS1101 Group1

Discussion 5

Lek Hsiang Hui

lekhsian @ comp.nus.edu.sg

http://www.comp.nus.edu.sg/~lekhsian/cs1101

Debugging in Dr Java

Checklist• Setting a breakpoint

– Before running program– During the debugging process

• Step into a method• Watches

Debugging in Dr Javapublic class Dummy{ public static void main(String[] args){ int tmp = 0; for (int i = 0 ; i < 10 ; i++){ tmp++; } dummyMethod(); } public static void dummyMethod(){ System.out.println("do nothing1"); System.out.println("do nothing2"); System.out.println("do nothing3"); }}

• Ctrl+B (toggle breakpoints at current line)• F7 (Resume/Continue Run)• F11 (Step Over)• F12 (Step Into)• Shift + F12 (Step Out)

Search And Sorting

Searching

• Linear Search– Go through all the items in an array

• Binary Search– Search by “cutting” the possible solutions by

½ each time

Sorting

• Bubble Sort

• Insertion Sort

• Selection Sort

Stable Sorts

• A stable sort is one where the relative ordering of elements with the same value is preserved after sorting.– Two elements having the same key appear in the same order in

the sorted sequence as they did in the original sequence.

• Example:– Stable sort:

Before sorting: 6 2 3a 8 3b 5 9

After sorting: 2 3a 3b 5 6 8 9

– Unstable sort:

Before sorting: 6 2 3a 8 3b 5 9

After sorting: 2 3b 3a 5 6 8 9

• Which of the three basic sorts – bubble sort, selection sort, insertion sort – is/are stable?

Slide taken from cs1101x lecture notes

Sorting Algorithms Comparison

Name Best Time Worst Time Stable(in general)

Bubble sort O(n) O(n2) Yes

Insertion sort O(n) O(n2) Yes

Selection sort

O(n2) O(n2) No

Sorting Algorithms Comparison

Name Best Time Worst Time Stable(in general)

Bubble sort about

n comparisons

about

n2 comparisons

Yes

Insertion sort about

n comparisons

about

n2 comparisons

Yes

Selection sort about

n2 comparisons

about

n2 comparisons

No

Sorting Question

Which of the 3 algorithms is best for data

that is almost sorted?

Recursion

Recursion

• Base case

• Recursive case

n! = n * (n-1) * (n-2) * … * 2 * 1 for n > 0 (recursive case)

0! = 1 (base case)

int fac(int n){

if (n == 0) return 1; else{ return n * fac(n-1); }

}

Recursion

• General approaches

• Think recursive– Split the problem into simple cases– Work in the opposite way

• Work from a base case• See what you need to do for the 2nd last case• etc

Recursion Practice

• Can you write a “while” method that does what the while loop is doing

Recursion Practice

• Can you write a “while” method that does what the while loop is doing

public static void While (int i){

if (cond){

//code of while body

While(i);

}

}

Recursion Practice

while(i > 0){ System.out.println(i); i--;}

public static void While (int i){ if (i > 0 ){ System.out.println(i); i--; While(i); } }

References

• Pass by Value– Java is always pass by value!

• Pass by Reference

Pass by Valuepublic static void main (String args[]){

//a is an Integer Object

//you can just think of it as a

//integer value

Integer a = new Integer(13);

addOne(a);

System.out.println(a);

}

public static void addOne(Integer number){

number = new Integer(number.intValue() + 1); System.out.println("number inside the method : " + number);

}

What do you think is the output?


//a is an Integer Object

//you can just think of it as a

//integer value


addOne(a);


}

public static void addOne(Integer number){

number = new Integer(number.intValue() + 1); System.out.println("number inside the method : " + number);

}

What do you think is the output?

number inside the method : 14

13



addOne(a);


}

public static void addOne(Integer number){…}

What is really happening on the JVM?

13

0x0010

The value of a is at memory location 0x0010

Note that this is really a simplified view of what’s happening



addOne(a);


}



13

0x0010

0x0010

a

0x0020

To find the value of a, just need to go to

memory location 0x0010This information is stored at

memory location 0x0020



addOne(a);


}



13

0x0010

0x0010 addOne(a)

What happens when you call a method is that another box with the value 0x0010 is passed in

(both are pointing to 0x0010)

What this means is that you can, only change the contents

at 0x0010, but you cannot define anew totally new object for a

(You are allowed to modify 0x0010)a

0x0020 0x0010

Copy of a

0x0030

Pass by Referencepublic static void main (String args[]){


addOne(a);


}



13

0x0010

0x0010 addOne(a)

In Pass by Reference,what happens call a method

is that the first box on the left is passed

to the method(0x0010)

What this means is that you can, change the contents

at 0x0010, and you can also give new pointer for Integer a

(You are allowed to modify both 0x0010, and 0x0020)

a

0x0020

Java (Pass by value)

• When you are passing primitive data types to methods (int, byte, double etc), you can think of it as you are giving the method the numerical value of the variable, not the variable itself


• When you are passing in references…– Objects, Arrays*– Can really think of Arrays are Objects

• You are given access to the content of that Object (but cannot modify the container)


int[] arr = new int[10];

method(arr);

public static void method(int[] a){

a = new int[20];

}

arrcontents

0x0010

0x0010

arr

0x0020 0x0010

Copy of arr

0x0030



method(arr);


a = new int[20];

}

arrcontents

0x0010

0x0010

arr

0x0020 0x0050

Copy of arr

0x0030

new array

When you comes out of the

method, you don’t really have the

previous box and 0x0050



method(arr);


a = new int[20];

}

arrcontents

0x0010

0x0010

arr

0x0020

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