30 December 2011

Java History

Java is a programming language expressly designed for use in the Standalone and Distributed environment of the Internet. Java was designed to have the "look and feel" of the C++ language, It is an OOPS-Object Oriented Programming model. Java can be used to create complete applications that may run on a single computer or be distributed among servers and clients in a network. It can also be used to build a small application module or Applet for use as part of a Web page. Applets make it possible for a Web page user to interact with the page.

Father of Java James Gosling Complete History

James Gosling Is The Father of the Java:

James A. Gosling, O.C., Ph.D. (born May 19, 1955 near Calgary, Alberta, Canada) is a famous software developer, best known as the father of the Java programming language.

 


Educational Qualifications of James Gosling:

           In 1977, James Gosling received a B.Sc in Computer Science from the University of Calgary. In 1983, he earned a Ph.D in Computer Science from Carnegie Mellon University, and his doctoral thesis was titled "The Algebraic Manipulation of Constraints". While working towards his doctorate, he wrote a version of emacs (gosmacs), and before joining Sun Microsystems he built a multi-processor version of Unix while at Carnegie Mellon University, as well as several compilers and mail systems.
Since 1984, Gosling has been with Sun Microsystems, and is generally known best as the founder of the Java programming language.
Thinking About Java


Implementations In Java:


          He is generally credited as the inventor of the Java programming language in 1991. He did the original design of Java and implemented its original compiler and virtual machine. For this achievement he was elected to the United States National Academy of Engineering. He has also made major contributions to several other software systems, such as NeWS and Gosling Emacs. He also cowrote the "bundle" program, a utility thoroughly detailed in Brian Kernighan and Rob Pike's book The Unix Programming Environment.


Java Always Belongs to 'SUCCESS'











More Details visit Wikipedia

Refer FAQ Questions and Improve Your Java Skills

Refer This Questions and Get the Job...!

1.What is the use of main method?
   Main() method is used to start the execution of the program.
   
2.What is overloading?
    Overloading is nothing but having the same name with different parameters.
   (or) The same function having different functionality.

3.What is overriding?
   Overriding is nothing but having same name and method signature.
   It is between  parent class and child class. Child class will always overrides parent class.

4.What are the access modifiers?
   Public, Private, Protected and Default.

5.What is Method Signature?
   Method signature is nothing but (return type of function + function name + no.of arguments).

6.What are the rules we have to follow in overloading?
   i)No.of parameters must be different.
   ii)The type of parameters must be different
   iii)The order of parameters
   iv) It will not depend on return type.

7.What are the Secondary Access Modifiers?
   Static, abstract, final, transient and volatile.

8.What is Shadowing?
   Declaring the same variables of parent class in the subclass is known as Shadowing.
   Example : super.x;  this.b;
  class ClassName {
    int  a; int  b;
    void setA(int a) {
      a = a; //here, the local variable has closer scope than the instance 
           // variable, so the expression set parameter equal to itself
      this.a = a; // this is the correct way to set the parameter to the 
            //instance variable.
    }
    void setB(int b) {
     this.b = b;
   }
 }
Let's look at another example,
 class ClassName {
    String name = "Instance Var."; 
    void someMethod() { 
       String name ="Local Var."; 
       System.out.println(name);   //Local Var.
       System.out.println(this.name); //Instance Var. 
    } 
  }
9.What are the types of data types in Java?
   byte, short, int, float, long, double, char and Boolean

10.Can we call the non static variables into static methods?
   No, vice versa is also not possible.

You cannot use a non-static variable in a static method. If this is what you are trying to do: 
class ClassName
{
 string someVariable;
 

 static void SomeMethod()
 {
  someVariable = "JavaHari";
 }
}
 You cannot do that. However, you can declare the variable static if you need to, like this: 
class ClassName
{
 static string someVariable;
 
 static void SomeMethod()
 {
  someVariable = "JavaHari";
 }
}
11.What is the  basic difference between the overloading and overriding?
   It is necessary to check the method signature in overloading, whereas
  it is not necessary in overriding. Method overloading between the same class methods and method overriding is between parent class method and child class method.

12.What is java bytecode?
   Bytecode is a highly optimized set of instructions designed to be executed by the java run time system, which is called the Java Virtual Machine (JVM)

13.What is java virtual machine?
   JVM is interpreter for the bytecode.

14.What is Encapsulation?
   Encapsulation is the mechanism that binds together code and the data it manipulates, and keeps the safe from outside interference and misuse.


15.What is Data Abstraction?
   An Abstraction is noting but keeping the data and methods in a single object.

16.What is inheritance?
   Inheritance is the process of getting [Base Class futures in to Derived Class] or [Data members & Methods of one class into another class ] is called Inheritance.
17.What are the types of inheritance?
   a) Single Level inheritance.
   b) Multiple inheritances
   c) Multi level inheritance
   d) Hybrid inheritance.
   e) Hierarchical inheritance.

18.Dose Java supports multiple inheritances?
   No.  But by using inner Interfaces it is possible.

19. What is polymorphism?
   Polymorphism is a feature that the process of representing one from to multiple forms is known as.

20. What is Type Costing?
   The conversion of one data type to another data type is known as
   Type Costing. It is possible in higher data type to lower. 

21. What is the range and width of long?
   Range is -9,223,372,036,854,775,808 to 9,223,372,036,854,775,807 and the width is 64.

22. What is the range and width of int?
   Range is -2,147,483,648 to  2,147,483,647 and width is 32.

23. What is the range and width of  short?
   Range is -32,768 to 32,767 and width is 16.

24. What is the range and width of byte?
   Range is -128 to 127 and width is 8.

25.  What is the declaration of an Array?
   int  Array[] = new int [5];

26. What is a class?
The process of binding data members and the associated methods in a single unit is known as.  
[OR]A class is a template for an object.

27. What is an Object?
   Object is instance of a class. 
   Instance is nothing but allocation sufficient memory space to Data members and Methods

28. What are called instance variables?
   The data or variables defined within a class are called instance variables.

29. What is a Constructor?
   Constructor is used to initialize the object. Constructors have no return type, not even void.
   For constructor the class name and method name must be same. 
30. What are the types of Constructors?
   Constructors are divided into 3 types : I) Default 2) Parameterized 3)Copy Constructor[object]

31. What are the types of explicit constructors?
   Default constructors and Parameterized constructors.

32. Where we use the key word 'this'?
   'this' is always a reference to the object on which the method was invoked.

33). Can we get the output using Constructors?
   No, These are used only for assigning values.

33. What is garbage Collection?
   If no reference to an object, that object is assumed to be no longer
   needed, and the memory occupied by the object can be reclaimed.
   This is known as garbage collection.

34. What is finalize method?
   Using finalize method we can define specific actions that will occur
   when an object is just about to be reclaimed by the garbage collector.

35. What is call by value and call by reference?
   When a simple type is passed to a method, it is done by use of call
   by-by-value. Objects are passed call by reference.

36. What is Recursion?
   It is a process of defining something in terms of itself.

37.  What is recursive?
   A method that calls itself is said to be recursive.

38. What is Static?
  Static is the key word, which is used to create a member that can be used by itself, without reference to a specific instance. It is a class level variable. It is illegal to refer to any instance variables inside of a static method.

39. Can we refer instance variables inside of a static method?
   No. We can't.
40.What is final?
   The final prevents its contents from being modified.

41.What is an inner class?
Inner class is a class, which is, defined in side another class.

42. What is an anonymous class?
Anonymous class is a class, which don't have any name.

43. What is super?
Super is used to get the properties of the previous class.

44. What is Dynamic Method Dispatch?
Dynamic method dispatch is the mechanism by which a call to an overridden
function is resolved at run time, rather than at compile time.

45. What is an abstract class?
Restriction of data is called Abstract. We can't create the instance.
These are virtual classes. It is same as interface but we can implement the methods.
 
46. What is wrapper class?
Wrapper classes are used to represent the primitive data types as objects.

47. Which key word is used to prevent Overriding?
Final

48. What is a package?
A package statement defines a name space in which classes are stored.

49. What is an interface?
It is a task for specific contract. It does not actually define
any implementations. It is not at all a class. It won't allow constructors.

50. What is public?
If we declare it as a public, classes, subclasses within the package
and outside the package, can access it.       

28 July 2011

C-Language Aptitude Questions & Answers

C-Aptitude Questions & Answers


1.      void main()
{
            int  const * p=5;
            printf("%d",++(*p));
}
Answer:
                        Compiler error: Cannot modify a constant value.
Explanation:   
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

2.      main()
{
            char s[ ]="man";
            int i;
            for(i=0;s[ i ];i++)
            printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
Answer:
                        mmmm
                       aaaa
                       nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally  array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the  case of  C  it is same as s[i].

3.      main()
{
            float me = 1.1;
            double you = 1.1;
            if(me==you)
printf("I love U");
else
                        printf("I hate U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value  represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) . 

4.      main()
            {
            static int var = 5;
            printf("%d ",var--);
            if(var)
                        main();
            }
Answer:
5 4 3 2 1
            Explanation:
When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.  

5.      main()
{
             int c[ ]={2.8,3.4,4,6.7,5};
             int j,*p=c,*q=c;
             for(j=0;j<5;j++) {
                        printf(" %d ",*c);
                        ++q;     }
             for(j=0;j<5;j++){
printf(" %d ",*p);
++p;     }
}

Answer:
                        2 2 2 2 2 2 3 4 6 5
            Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.
           
6.      main()
{
            extern int i;
            i=20;
printf("%d",i);
}

Answer: 
Linker Error : Undefined symbol '_i'
Explanation:
                        extern storage class in the following declaration,
                                    extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

7.      main()
{
            int i=-1,j=-1,k=0,l=2,m;
            m=i++&&j++&&k++||l++;
            printf("%d %d %d %d %d",i,j,k,l,m);
}
Answer:
                        0 0 1 3 1
Explanation :
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression  i++ && j++ && k++’ is executed first. The result of this expression is 0    (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

8.      main()
{
            char *p;
            printf("%d %d ",sizeof(*p),sizeof(p));
}

Answer:
                        1 2
Explanation:
The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

9.      main()
{
            int i=3;
            switch(i)
             {
                default:printf("zero");
                case 1: printf("one");
                           break;
               case 2:printf("two");
                          break;
              case 3: printf("three");
                          break;
              } 
}
Answer :
three
Explanation :
The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

10.  main()
{
              printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

11.  main()
{
            char string[]="Hello World";
            display(string);
}
void display(char *string)
{
            printf("%s",string);
}
            Answer:
Compiler Error : Type mismatch in redeclaration of function display
            Explanation :
In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

12.  main()
{
            int c=- -2;
            printf("c=%d",c);
}
Answer:
                                    c=2;
            Explanation:
Here unary minus (or negation) operator is used twice. Same maths  rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can  only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.

13.  #define int char
main()
{
            int i=65;
            printf("sizeof(i)=%d",sizeof(i));
}
Answer:
                        sizeof(i)=1
Explanation:
Since the #define replaces the string  int by the macro char

14.  main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}
Answer:
i=0


            Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol.  ! is a unary logical operator. !i (!10) is 0 (not of true is false).  0>14 is false (zero).

15.  #include<stdio.h>
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
77       
Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
 Now performing (11 + 98 – 32), we get 77("M");
 So we get the output 77 :: "M" (Ascii is 77).

16.  #include<stdio.h>
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8}  };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}
Answer:
SomeGarbageValue---1
Explanation:
p=&a[2][2][2]  you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.
           
17.  #include<stdio.h>
main()
{
struct xx
{
      int x=3;
      char name[]="hello";
 };
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
            Answer:
Compiler Error
Explanation:
You should not initialize variables in declaration

18.  #include<stdio.h>
main()
{
struct xx
{
int x;
struct yy
{
char s;
            struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

19.  main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}
Answer:
hai
Explanation:
\n  - newline
\b  - backspace
\r  - linefeed

20.  main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:
45545
Explanation:
The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the  evaluation is from right to left, hence the result.

21.  #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}
Answer:
64
Explanation:
the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64
 
22.  main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s   %s",p,p1);
}
Answer:
ibj!gsjfoet
            Explanation:
                        ++*p++ will be parse in the given order
Ø  *p that is value at the location currently pointed by p will be taken
Ø  ++*p the retrieved value will be incremented
Ø  when ; is encountered the location will be incremented that is p++ will be executed
Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print anything.

23.  #include <stdio.h>
#define a 10
main()
{
#define a 50
printf("%d",a);
}
Answer:
50
Explanation:
The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

24.  #define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}
Answer:
100
Explanation:
Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input  program to compiler looks like this :
                        main()
                        {
                             100;
                             printf("%d\n",100);
                        }
            Note:  
100; is an executable statement but with no action. So it doesn't give any problem

25.  main()
{
printf("%p",main);
}
Answer:
                        Some address will be printed.
Explanation:
            Function names are just addresses (just like array names are addresses).
main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.

27)       main()
{
clrscr();
}
clrscr();
           
Answer:
No output/error
Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).

28)       enum colors {BLACK,BLUE,GREEN}
 main()
{
 
 printf("%d..%d..%d",BLACK,BLUE,GREEN);
  
 return(1);
}
Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not explicitly defined.

29)       void main()
{
 char far *farther,*farthest;
 
 printf("%d..%d",sizeof(farther),sizeof(farthest));
  
 }
Answer:
4..2 
Explanation:
            the second pointer is of char type and not a far pointer

30)       main()
{
 int i=400,j=300;
 printf("%d..%d");
}
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program,then printf will take garbage values.

31)       main()
{
 char *p;
 p="Hello";
 printf("%c\n",*&*p);
}
Answer:
H
Explanation:
* is a dereference operator & is a reference  operator. They can be    applied any number of times provided it is meaningful. Here  p points to  the first character in the string "Hello". *p dereferences it and so its value is H. Again  & references it to an address and * dereferences it to the value H.

32)       main()
{
    int i=1;
    while (i<=5)
    {
       printf("%d",i);
       if (i>2)
              goto here;
       i++;
    }
}
fun()
{
   here:
     printf("PP");
}
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is available in function fun() Hence it is not visible in function main.

33)       main()
{
   static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
    int i;
    char *t;
    t=names[3];
    names[3]=names[4];
    names[4]=t;
    for (i=0;i<=4;i++)
            printf("%s",names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.

34)       void main()
{
            int i=5;
            printf("%d",i++ + ++i);
}
Answer:
Output Cannot be predicted  exactly.
Explanation:
Side effects are involved in the evaluation of   i

35)       void main()
{
            int i=5;
            printf("%d",i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.
  
36)       #include<stdio.h>
main()
{
int i=1,j=2;
switch(i)
 {
 case 1:  printf("GOOD");
                break;
 case j:  printf("BAD");
               break;
 }
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).
            Note:
Enumerated types can be used in case statements.

37)       main()
{
int i;
printf("%d",scanf("%d",&i));  // value 10 is given as input here
}
Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0.  Here 10 is given as input which should have been scanned successfully. So number of items read is 1.

38)       #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
            }
Answer:
100

39)       main()
{
int i=0;

for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer:
            1
Explanation:
before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

40)       #include<stdio.h>
main()
{
  char s[]={'a','b','c','\n','c','\0'};
  char *p,*str,*str1;
  p=&s[3];
  str=p;
  str1=s;
  printf("%d",++*p + ++*str1-32);
}
Answer:
M
Explanation:
p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from 32.
i.e. (11+98-32)=77("M");
           
41)       #include<stdio.h>
main()
{
  struct xx
   {
      int x=3;
      char name[]="hello";
   };
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the structure declaration

42)       #include<stdio.h>
main()
{
struct xx
 {
              int x;
              struct yy
               {
                 char s;
                 struct xx *p;
               };
                         struct yy *q;
            };
            }
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared.

43)       main()
{
 extern int i;
 i=20;
 printf("%d",sizeof(i));
}
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn't find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.

44)       main()
{
printf("%d", out);
}
int out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.

45)       main()
{
 extern out;
 printf("%d", out);
}
 int out=100;
Answer:
100     
            Explanation:  
This is the correct way of writing the previous program.
                 
46)       main()
{
 show();
}
void show()
{
 printf("I'm the greatest");
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().

47)       main( )
{
  int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
  printf(“%u %u %u %d \n”,a,*a,**a,***a);
        printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);
       }
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
                  The given array is a 3-D one. It can also be viewed as a 1-D array.
                                                                                                                                                                                                                                                                     
2
4
7
8
3
4
2
2
2
3
3
4
   100  102  104  106 108   110  112  114  116   118   120   122

thus, for the first printf statement a, *a, **a  give address of  first element . since the indirection ***a gives the value. Hence, the first line of the output.
for the second printf a+1 increases in the third dimension thus points to value at 114, *a+1 increments in second dimension thus points to 104, **a +1 increments the first dimension thus points to 102 and ***a+1 first gets the value at first location and then increments it by 1. Hence, the output.

48)       main( )
{
  int a[ ] = {10,20,30,40,50},j,*p;
  for(j=0; j<5; j++)
    {
printf(“%d” ,*a);
a++;
    }
    p = a;
   for(j=0; j<5; j++)
      {
printf(“%d ” ,*p);
p++;
      }
 }
Answer:
Compiler error: lvalue required.
                       
Explanation:
Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

49)       main( )
{
 static int  a[ ]   = {0,1,2,3,4};
 int  *p[ ] = {a,a+1,a+2,a+3,a+4};
 int  **ptr =  p;
 ptr++;
 printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
 *ptr++;
 printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
 *++ptr;
 printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
 ++*ptr;
       printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);
}
Answer:
            111
            222
            333
            344
Explanation:
Let us consider the array and the two pointers with some address
a   
0
1
2
3
4
   100      102      104      106      108
                                                           p
100
102
104
106
108
                                       1000    1002    1004    1006    1008
           ptr 
1000
2000
After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling factor) = 1,  *ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a value 102  so the value is (102 – 100)/(scaling factor) = 1,  **ptr is the value stored in the location pointed by  the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is  1, 1, 1.
After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.
After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.
After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.

50)       main( )
{
 char  *q;
 int  j;
 for (j=0; j<3; j++) scanf(“%s” ,(q+j));
 for (j=0; j<3; j++) printf(“%c” ,*(q+j));
 for (j=0; j<3; j++) printf(“%s” ,(q+j));
}
Explanation:
Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1. Suppose the inputs are MOUSE,  TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as

M

O

U

S
E
\0
When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.

M

T

R

A

C

K

\0

The third input  starts filling from the location 102

M

T

V

I

R

T

U

A

L

\0
This is the final value stored .
The first printf prints the values at the position q, q+1 and q+2  = M T V
The second printf prints three strings starting from locations q, q+1, q+2
 i.e  MTVIRTUAL, TVIRTUAL and VIRTUAL.
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