Oracle Placement Paper -: SQL Questions With Answers

Oracle Previous Years Solved Sample Placement Papers

1. What is the output of the program:
   

   void main()
   {
       int j[10] = {9, 7, 5, 3, 1, 2, 4, 6, 9};
       int i = 1;
       clrscr();
       for(; i < 9; i++)
           printf("%d ", --j[i++]);
       getch();
   }
           

   A. 6,2,1,5

   B. 6,2,1,5,7

   C. Error Message

   D. core dump Answer: Option A

2. Solution: The loop iterates and decrements the values in the array based on the given logic, producing the sequence 6, 2, 1, 5.
3. What is the output of the program:
   

   void main()
   {
       int i, j, k, n = 5;
       clrscr();
       for(i = 5; i > 0; i--)
       {
           j = 1 < i;
           k = n & j;
           k == 0 ? printf("0") : printf("1");
       }
       getch();
   }
           

   A. 00011

   B. 11110 Answer: Option B

   C. 11001

   D. 11100

4. Solution: The program outputs binary based on the bitwise AND operation and the loop conditions.
5. Which of the following storage class(es) became the global variable for the entire program:
   

   A. Extern Answer: Option A

   B. Static

   C. Auto

   D. Register

6. Solution: The `extern` keyword makes a variable global and accessible across different files.
7. What is the output of the program, if integer occupies 2 bytes memory:
   

   union
   {
       int a;
       char b;
       char c[10];
   } u1;
   
   void main()
   {
       int l = sizeof(u1);
       printf("%d", l);
       getch();
   }
           

   A. 13

   B. 10 Answer: Option B

   C. 16

   D. None of the above

8. Solution: The union size is determined by the largest member, which is the char array of size 10.
9. What is the output of the program:
   

   void main()
   {
       fork();
       printf(" Hello World");
       getch();
   }
           

   A. Hello World

   B. Hello World Hello World Answer: Option B

   C. Error Message

   D. None of these

10. Solution: The `fork()` system call creates a new process, resulting in the message being printed twice.
11. What is the output of the program:
    

    void main()
    {
        struct a
        {
            int i;
            char *st1;
        };
        typedef struct a ST;
        ST *str1;
        str1 = (ST*)malloc(100);
        str1->i = 100;
        strcpy(str1->st1, "Welcome to Oracle");
        printf(" %d%s\n", str1->i, str1->st1);
        getch();
    }
            

    A. core dump

    B. will not compile

    C. 100, Welcome to Oracle Answer: Option C

    D. None of these

12. Solution: Memory is allocated, and the values are printed as expected.
13. What is the output of the program:
    

    void main()
    {
        int i, j, k;
        i = 2;
        j = 4;
        k = i++ > j & 2;
        printf("%d\n", k);
        if(++k && ++i < --j || i++)
        {
            j = ++k;
        }
        printf(" %d %d %d", i, -j--, k);
        getch();
    }
            

    A. 4, -3, 2

    B. 5, -3, 2

    C. 4, -2, 2

    D. 5, -2, 2 Answer: Option D

14. Solution: The operations and precedence rules determine the final output.
15. Which of the following is not true in case of Command line arguments:
    

    A. The argc parameter is used to hold the number of arguments in the command line and is an integer

    B. The argv parameter is a pointer to an array of character pointers and each one points to command line arguments

    C. The argv[1] always points to program name Answer: Option C

    D. None of the above

16. Solution: The program name is always `argv[0]`, not `argv[1]`.
17. What is the output of the program:
    

    #include
    #include
    #define sqr(a) a*a
    
    void main()
    {
        int a = 10, b = 1, c;
        c = sqr(10+1);
        printf("Sqr Root of (10+1) is %d", c);
        getch();
    }
            

    A. 121

    B. 21 Answer: Option B

    C. 13

    D. Syntax Error

18. Solution: Macro substitution leads to incorrect evaluation of the expression.
19. What is the output of the program:
    

    #include
    #include
    
    void main()
    {
        int i, j = 20;
        clrscr();
        for(i = 1; i < 3; i++)
        {
            printf("%d,", i);
            continue;
            printf("%d", j);
            break;
        }
        getch();
    }
            

    A. 1,20

    B. 1,20,1,20

    C. 1,2 Answer: Option C

    D. 1,2,20,20

20. Solution: The `continue` statement skips the rest of the loop body, so only `i` is printed.