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Design is not just what it looks like and feels like. Design is how it works." - Steve Jobs
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Hello +Kush Jain
Java does not support typedefs, defines, or a preprocessor. The declaration of named constants is supported in Java through use of the final keyword.

- Java supports classes, but does not support structures or unions.

- Global functions and global data are not allowed in Java.

- All classes in Java ultimately inherit from the Object class. C++ can have inheritance trees that are completely unrelated to one another.

- The interface concept is not supported by C++.

- Java does not support multiple inheritance.

- Java does not support the goto statement. However, it does support labeled break and continue statements, a feature not supported by C++.

- Java does not support operator overloading.

- Java does not support automatic type conversions (except where guaranteed safe).

- Unlike C++, Java has a String type, and objects of this type are immutable.

- Java does not support pointers

- The scope resolution operator (::) like in C++ is not used in Java.

- The size of each primitive type is the same regardless of the platform.

- There is no unsigned integer type in Java.

- C++ requires that classes and functions be declared before they are used. This is not necessary in Java.

- There are no destructors in Java.

- Unlike C++, Java has built-in support for program documentation.

- Java is more robust than C++.


Are there any inbuilt functions in c that help clear the screen?

Hello sir,
I am new to programming and I would like to know the difference between C++ and Java (apart from Syntax)... #Kossinedoubt

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Please check the below links for more details :

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Hello +Janvi Dholakia,
Malloc() allocates a memory block of the given size and returns the pointer to the beginning of the block. It doesn't initialize the memory block i.e the memory block will contain garbage values.
Calloc() allocates a memory block as well as initializes memory to zero.

What is the difference between calloc() and malloc() in C? #Kossinedoubt

Hi +Pratik Bhambhani,
In the case of a System.exit(), No. The call to the exit method never returns, and it doesn't throw an exception either. Therefore the "enclosing" finally clauses for the thread never get executed. This is a very popular tricky Java question and its tricky because many programmer think that finally block always executed. This question challenge that concept by putting return statement in try or catch block or calling System.exit from try or catch block. Answer of this tricky question in Java is that finally block will execute even if you put return statement in try block or catch block but finally block won't run if you call System.exit from try or catch.

Jude Sir's teaching methodology is superb. The way he simplifies concepts is exemplary. He addresses every doubt in the class and motivates students constantly. Programming has become interesting for me because of him.

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For all those who have the same doubt as +Rosanlal Behera
Here is an extensive coverage of the segmentation fault:
Troubleshooting Segmentation Violations/Faults

A common run-time error for C programs by beginners is a "segmentation violation" or "segmentation fault." When you run your program and the system reports a "segmentation violation," it means your program has attempted to access an area of memory that it is not allowed to access. In other words, it attempted to stomp on memory ground that is beyond the limits that the operating system (e.g., Unix) has allocated for your program.
Any time your program gives a "segmentation violation" or "segmentation fault" error, review this document for tips on correcting the error.

Common causes of this problem:

Improper format control string in printf or scanf statements:
Make sure the format control string has the same number of conversion specifiers (%'s) as the printf or scanf has arguments to be printed or read, respectively, and that the specifiers match the type of variable to be printed or read. This also applies to fprintf and fscanf.
Forgetting to use "&" on the arguments to scanf:
Function scanf takes as arguments the format control string and the addresses of variables in which it will place the data that it reads in. The "&" (address of) operator is used to supply the address of a variable. It is common to forget to use "&" with each variable in a scanf call. Omitting the "&" can cause a segmentation violation.
Accessing beyond the bounds of an array:
Make sure that you have not violated the bounds of any array you are using; i.e., you have not subscripted the array with a value less than the index of its lowest element or greater than the index of its highest element.
Failure to initialize a pointer before accessing it:
A pointer variable must be assigned a valid address (i.e., appear on the left-hand-side of an assignment) before being accessed (i.e., appearing on the right-hand-side of an assignment). Make sure that you have initialized all pointers to point to a valid area of memory. Proper pointer initialization can be done several ways. Examples are listed below.
Incorrect use of the "&" (address of) and "*" (dereferencing) operators:
Make sure you understand how these operators work. Know when they should be applied and when not to apply them. As mentioned above, it is common to forget to use "&" with each variable in a scanf call. Remember, scanf requires the address of the variables it is reading in. Especially, know when "&" and "*" are absolutely necessary and when it is better to avoid using them.
Proper pointer initialization:

One common way is to assign the pointer an address to a previously defined variable. For example:

int * ptr;
int variable;

ptr = &variable;
Or, equivalently,

int variable;
int ptr = &variable;
Other common ways include assigning the pointer the address of memory allocated with matrix, vector, calloc, or malloc or other equivalent allocation functions. Remember, a pointer must be initialized to a value (i.e., assigned a value by appearing on the left-hand-side of an assignment statement) BEFORE you attempt to access it!
Minimizing the use of pointer variables.

Also, many times a function requires that an address (corresponding to a parameter of pointer type) be sent to it as an argument (as is true of many of the Numerical Recipes in C functions). The standard function scanf is an example of such a function.
In these cases, it is usually best to simply declare a variable of the correct type before calling the function and just sending the address of the variable to the function. In fact, that is what is intended in the vast majority of these cases. And, that's how you usually scanf:

double x_initial; / initial guess /
scanf("%lf",&x_initial); / Read the initial guess. /
For example, see how 'idum' is used below:

long idum = -1; / initialize idum to be a negative integer /

/ generate a random number from the normal distn./
x = normal(&idum,average,stddev);
The function normal expects an address to a variable of type long. That's what we send it without explicitly using a pointer variable in the calling routine.
Troubleshooting the problem:

Check EVERY place in your program that uses pointers, subscripts an array, or uses the address operator (&) and the dereferencing operator (). Each is a candidate for being the cause of a segmentation violation. Make sure that you understand the use of pointers and the related operators.
If the program uses many pointers and has many occurrences of & and *, then add some printf statements to pinpoint the place at which the program causes the error and investigate the pointers and variables involved in that statement.

Remember that printf statements for debugging purposes should have a new-line character (\n) at the end of their format control strings to force flushing of the print buffer.
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