Programming Questions

A is incorrect. It is acceptable to use assertions to test the arguments of private methods.

B is incorrect. While assertion errors can be caught, Sun discourages you from doing so.

struct emp

{

     char n[20];

      int age;

};

main()

{

    struct emp e1 = {"Dravid", 23};

   struct emp e2;

   scanf ("%s %d",e2.n, & e2.age);

   if( structcmp (e1,e2) ==0)

        printf ("The structures are equal");

   else

         printf ("The structures are unequal");

}

structcmp ( struct emp x, struct emp y)

{

     if (strcmp (x.n,y.n) ==0)

           if (x.age == y.age)

            return (0);

            return (1);

}

In short, if you nee to compare two structures, you'll have to write your own function to do so which carries out the comparison field by field.

When a function accepts a variable number of arguments , its prototype cannot provide any information about the number of arguments and type of those variable arguments. Hence the compiler cannot warn about the mismatches. The programmer must make sure that arguments match or must manually insert explicit typecast.

Till the stack doesn't overflow

Once a method declared as Final cannot be  overriden

#include "mem.h"

#include "alloc.h"

main()

{

      int area;

      char *dest;

      char src[] = "Life is the camera and you are the target"

                                  "so keep smiling always";

      area = sizeof (src);

      dest = malloc (area);

      memmove (dest, src, area);

      printf("\n%s", dest);

      printf("\n%s",src);

}

Output: 12

Explanation:
Binary value of a=1 is 00001 (in 5 bit)
Binary value of b=3 is 00011 (in 5 bit)
Binary value of c=3 is 000011 (in 6 bit)

In memory it is represented as:
Let address of bit1 is 500 which initialize to char pointer p. Since can is one byte data type so p++ will be 501. *p means content of memory location 501 which is (00001100) and its binary equivalent is 12. Hence output is 12.