tasklist - cmd in windows similar to ps cmd in linux

The people who have worked in Unix and its flavours, would have definitely come across a command called 'ps' (which stands for 'process status'). As such the 'ps' command gives you the list of processes running and the pid (process id) etc.,

There is a command to achieve the same in Windows OS too. That is 'tasklist'. Just execute in the command prompt. Here it gives the "Image Name", "PID", "Session name", "Session#", "Mem Usage".

Image Name -> the name of the process running
PID -> as you know, the process id (unique id which the OS refers to)
Session name -> it shows 'Console' but have to check what it really means
Session # -> should be a session id and for console it gives '0'.
Mem usage -> the amount of memory each task/process occupies.

Here is the sample screenshot of the same.

It looks like the CUI version of the 'task manager' :)

NOTE: It may not work in all versions of Windows. As such i have tested in Windows XP. I believe it must definitely work in 2000 also. Not sure about the previous versions of 98,95 etc., will check and update later.

Pass by Reference for object references holds good really?

Pre-requisite: Well, If you want to refreshwith the concepts of "Pass-by-value Vs Pass-by-reference", you may please refer this previous entry in this blog.

With respect to Java, it "passes everything by value". I repeat, "everything" by value only.

Generally, we may think that passing by values will hold good only for primitive types and values but not for the object references.

This everything includes even the references to objects.

Lets say, you have a method which takes an object as a parameter as follows


public void doChange(MyClass myClassObj)
{
//do something..
}


And you call that method from some other piece of code as follows

//calling method
public void method1()
{
MyClass myClassObj = new MyClass();
doChange(myClassObj);
}

The reference variable myClassObj is being assigned to an object of class MyClass in Heap. Just like primitive variables holding bit patterns to represent a value, reference variables also store the bit patterns to reach an object.

Once you call the method doChange() by passing the reference variable "myClassObj" to it, the value of the bit patterns held by "myClassObj" is being copied and passed into the called method doChange(). In the called method doChange(), the copied-and-sent bit pattern is received in the same name as that of the original reference variable "myClassObj". But the compiler treats that as a different one just like primitives.

Lets say the received reference variable as "myClassObjLocal" for easy understanding. In this stage, both the original refernece variable "myClassObj" and the received reference variable inside the doChange() method "myClassObjLocal" (to compiler) point to the same object in Heap.

Now if you change the state of the object being pointed by, it will reflect to both the reference variables because both of them point to the same object in Heap. Whereas,if you change the received reference variable "myClassObjLocal" to point to a new object (reassign a different object), it does NOT reflect back to the original object because the bit pattern to the "myClassObjLocal" alone gets changed!

If you are clear with the above paragraphs, lets go to an example to make it more clear.

Lets take a small example.

class TestObjRef
{
int intValue;

public TestObjRef()
{
intValue = 1;
}

public static void changeStateOfObject(TestObjRef obj)
{
obj.intValue = 2;
}

public static void changeReference(TestObjRef obj)
{
obj = new TestObjRef();
obj.intValue = 9;
}

public static void main(String[] args)
{
TestObjRef obj1 = new TestObjRef();
System.out.println("obj1.intValue (1) = "+obj1.intValue);
changeStateOfObject(obj1);
System.out.println("obj1.intValue (2) = "+obj1.intValue);
changeReference(obj1);
System.out.println("obj1.intValue (3) = "+obj1.intValue);
}
}


Running the above code produces the following output:

Output:
obj1.intValue (1) = 1
obj1.intValue (2) = 2
obj1.intValue (3) = 2


changeStateOfObject Method

This is because, changeStateOfObject() method just changes the value of the variable "intValue" which definitely constitutes the state of object. And as such, both the "obj1" in main() method and "obj" in changeStateOfObject() method point to the same object. Means, they both hold the same bit patterns to reach a single object of TestObjRef class in heap.

That's why the output in the second line shows the changed state of object "2" as the value of "intValue" property.

changeReference() Method

If you look at the changeReference() method, by the time of receiving the argument, both the "obj1" reference variable in main() method and received reference variable "obj" in changeReference() method both hold the same bit patterns to reach the same and single TestObjRef class in Heap.

But inside the method, the received "obj" reference variable is reassigned to a newly created object of TestObjRef class. In this case, only the bit pattern of the "obj" reference variable inside the changeReference() method is changed and the change is NOT reflected back to the originally sent reference variable "obj1".

That's why the third line in the output still shows the value of "intValue" as "2" since the changed value "9" is only reflected in the received local reference variable "obj" in changeReference() method.


Note :
The received argument (object reference variable) is treated local to the method. That means, the scope of the variable is only local to the called method just like primitives and the variable cannot be accessed outside the method.

Pass by value Vs Pass by Reference

We all would have read the two terms "pass-by-value" and "pass-by-reference" especially when you deal with methods (which are also called as functions).

Let's just have a quick glance at both the terms with a small example.

The definition of these terms goes as follows.

Pass-by-value: A copy of the value is being passed from the calling method to the called method.

What exactly do you mean by copy of the value? In general, we tend to pass the variable for the value to be passed. In such case, whatever the value the variable holds is being copied and passed across. That means, you are taking a xerox of the o riginal value and pass the xeroxed copy to the called method.

What are the effects?
Since you have only passed the xeroxed copy and NOT the original value, whatever changes you do in the called method, the change is not reflected back to the original variable.

A Sample program for pass-by-value:

void changeValue(int a)
{
int b = ++a;
printf("changeValue -> a, b contain: %d,%d",a,b);
}

int changeAndReturn(int a)
{
int b = ++a;
printf("changeAndReturn -> a,b contain : %d,%d",a,b);
return b;
}

int main()
{
int i = 1;
changeValue(i);
j = i;
printf("\ni and j is : %d,%d\n",i,j);
printf("\ncalling changeAndReturn()..");
changeAndReturn(i);
j = i;
printf("\nNow i and j contain: %d,%d",i,j);
}

Output:
changeValue-> a, b contain: 2 2
i and j is: 1,1
calling changeAndReturn()..
changeAndReturn -> a,b contain: 2 2
Now i and j contain: 1 2

Explanation:
Let's see how exactly the values are dealt. In the main() method, you are calling changeValue() method as

changeValue(i);

At this time, the value of i (which is 1) is being copied (Remember! xeroxed) and the xeroxed copy is what getting passed to the changeValue() method. Inside the changeValue() method, you get (receive) the passed value in the variable name "a". Note: Even if you have received the passed argument as the same name "i" in the function, the compiler would treat both of them differently!

Soon after you take a xerox copy of the value of the variable and use it for calling functions, the connection between the copied value and the original value is cut. There is no relation between those two values and /or variables.

Inside the method changeValue(), we do the following steps:

(i) receive the value passed in the variable whose name is 'a' (implicit)
(ii) declare an int variable named 'b'
(iii) increment the received value in its variable 'a' and assign the incremented value to 'b'

Since because the original value ('i' in main) does NOT hold any relationship with its xeroxcopy ('a' here), the changes you do in the called function (incrementing the value in changeValue()) does NOT reflect back.

That's why in the main after the function call returned, the printf statement prints the value both i and j contain 0. Once the function call is returned, the present focus is in main() function and now we are assigning the value of 'i' to 'j' which is 0 only. This clearly shows that there is NO effect on the variable 'i' even after calling the function changeValue() which does some change to it actually.

Just to have an effect, we have one more method called changeAndReturn() which changes the value and returns it. In the calling function (main()), we get the returned value and assign it to 'j'. That's why the second set of output shows the incremented value '2' for both 'i' and 'j'.

Pass-by-reference: The memory address of the variable is being passed instead of making a copy of the value. This holds good for C, C++ languages (as far as i know as such).

What are the effects?
Since you pass the memory address itself to the function and you obtain the value from the received memory address inside the called function, you get the full control over the value. That means, whatever change you do, it will be reflecting in the original variable in the calling function eventhough you do NOT return either the memory address or value back.

It is because, you have played around with the memory directly instead of values and in both the calling and called functions, the values are obtained and manipulated with respect to memory only. So the change is reflected back.

Let's see a sample program for pass-by-reference:

Sample Program for pass-by-reference:

void changeValue(int *int_ptr)
{
//doing an increment operation on the 'dereferenced' pointer
int b = ++(*int_ptr);
printf("changeValue -> a, b contain: %d,%d",a,b);
}

int main()
{
int i = 1;
int *int_ptr = &i;
printf("\ncalling changeValue()..");
//Note: we are passing the pointer to the function! (the 'address')
changeValue(int_ptr);
//assigning the value of 'i' to 'j'
int j = i;
printf("\ni and j is : %d,%d\n",i,j);
}

Output:
changeValue-> a, b contain: 2 2
calling changeValue()..
i and j is: 2,2

Explanation:

If you carefully look at the implementation and way of calling the function changeValue(), it actually receives a pointer to an integer! that means, we are passing the 'memory address' of an integer variable.

Inside the function, we do the following steps:

(i) receive the address of the integer variable in the name 'int_ptr' (the same name as in the calling function, here main(). But remember! for the compiler, they both are different!)
(ii) we declare an integer variable 'b'
(iii) we dereference the pointer variable ('dereferencing' means obtaining the value present in the location where the pointer is pointing to)
(iv) we increment the value being obtained by the dereferencing operation
Note: here the value is incremented directly in the memory location
(v) and assign the incremented value to the variable 'b'

One important thing is that, we do NOT return any value back to the calling function. But the changes happend in the memory location directly in the called function (changeValue()). Once the control comes back to the calling function (main()), we are still accessing the value in the same memory location through the original variable 'i' and assigning to 'j'.

That's why we see both i and j having the value '2'.

How to get the list of files in a directory and save it in a file

You may need to get the list of files inside a folder with the appropriate structures being maintained.

This tip will help you to get the complete list of filenames within a directory structure in Windows OS.

(1) Go to Command Prompt
(2) Navigate to any directory of your interest
(3) Type dir /b /s "Root-directory" > "filename.extension"

whereas

/b -> use bare format (no heading information or summary)
{summary is the date time info, size , file or directory attribute etc.}

/s -> displays file in specified directory and all subdirectories.

Root-directory -> is the directory from which you want the file list info
{if Root-directory is omitted it will process from the current directory}

> -> Redirection Operator or Symbol.
{the output of the command on its left would be sent to its right}

fileName.extension -> any file name with extension you want
{example. directory-structure.txt}

Make it Automated in a Batch File

You can even write a batch file with another line to open it in the
appropriate application.

dir /b /s "Root-directory" > "filename.extension"
notepad "filename.extension"

Save these two lines in a batchfile as ".bat" and run this batch file from command prompt. It would automatically save it in the file and gets opened in Notepad. You could use any other convenient applications to open as you wish.

Hope you may need not take screenshots for directory structure {if at all you had been doing so} ;-).

Tip to increase the efficiency while deleting huge volume of rows

This tip would be very much useful when you have enormous rows in a table and you are trying to delete/remove the rows using the usual, "DELETE FROM <tableName>;" statement.

You may ask whats the consequences of using the DELETE statement and thats what its purpose is. But, not just deletion, for that matter any DML statement will make use of something called TRANSACTION LOG which is nothing but the buffer/temporary storage where all the data would be kept until there is an explicit COMMIT or ROLLBACK.

Once a Commit is issued, the data kept in the Transaction Log is written into the database files in its own appropriate format (.dat for data files, .inx for index files in IBM DB2 database - which may vary in Oracle, MySQL, SQL Server etc..).

When a rollback is issued, the data in the temporary Transaction Log is removed and thereby no state change happens to the data files.

Moreover, its gonna be time consuming wherein the time is directly proportional to the dense of data.

When you deal with large number of data to be dealt with, be it inserting, updating or deleting, you might end up in something called, "The Transaction Log is FULL" while doing the operation. Most of us might not have encountered this situtation as we do just deal with 100s or max 1000s of data. People who are dealing with lakhs, millions of rows might have been aware of this. Say, who are in Data Warehousing, Business Objects etc.

When we encountered the situation, one of my teammates in my client place (Yadati Ravishankar) had come up with a query what he found out in googling. This query does the magic by just clearing up the table with very very less time compared to that of plain old DELETE statement.

Here you go...

"ALTER TABLE [<SchemaName>.]<tablename> ACTIVATE NOT LOGGED INITIALLY WITH EMPTY TABLE;"

This query does alter the table with empty rows but while doing the same it requests the db engine not to activate the Logging. Thereby it is able to get the request serviced within seconds!!

Try with huge data buddies..then only you can make out the difference!!

Good luck!

Java #Q1: What would be the output?

public class Test
{
     public void method1(Object o)
     {
          System.out.print("object");
     }
    public void method1(String s)
    {
          System.out.print("string");
    }
    public static void main(String[] args)
    {
           Test t = new Test();
           t.method1(null);
    }
}

When running the program, the output would be
(a) object
(b) string
(c) objectstring
(d) stringobject


Answer:
(b) string

Explanation:
It is the hierarchy of objects being searched for a match, which is quite similar to the exception objects of catch blocks. The search process is done in the bottom-up manner wherein the object of class of least node is searched first and its continued upwards to the super/parent classes till the end is reached.