C Puzzles : Questions and Answers - Level 2

 C Puzzles : Questions and Answers - Level 2

Questions
L2.Q1 : Write the output of this program

#include

main()
{
    int *a, *s, i;

    s = a = (int *) malloc( 4 * sizeof(int));

    for (i=0; i<4; i++) *(a+i) = i * 10;

    printf("%d\n", *s++);
    printf("%d\n", (*s)++);
    printf("%d\n", *s);
    printf("%d\n", *++s);
    printf("%d\n", ++*s);
}

Solution for L2.Q1
L2.Q2 : Checkout this program result

#include

void fn(int);

static int  val = 5;

main()
{
    while (val --) fn(val);
    printf("%d\n", val);
}

void fn(int val)
{
    static int val = 0;

    for (; val < 5; val ++) printf("%d\n", val);
}

Solution for L2.Q2
L2.Q3 : Can you predict the output of this program ?

#include

main()
{
    typedef union {
        int     a;
        char    b[10];
        float    c;
    } Union;

    Union    x, y = { 100 };

    x.a = 50;
    strcpy (x.b, "hello");
    x.c = 21.50;

    printf ("Union 2 : %d %s %f\n", x.a, x.b, x.c);
    printf ("Union Y : %d %s %f\n", y.a, y.b, y.c);
}

Solution for L2.Q3
L2.Q4 : Print the output of the program

#include

main()
{
    struct Data {
        int     a;
        int    b;
    } y[4] = { 1, 10, 3, 30, 2, 20, 4, 40};

    struct Data *x = y;
    int i;

    for(i=0; i<4; i++) {
        x->a = x->b, ++x++->b; 
        printf("%d %d\t", y[i].a, y[i].b);
    }
}

Solution for L2.Q4
L2.Q5 : Write the output of this program

#include

main()
{
    typedef struct {
        int    a;
        int     b;
        int     c;
        char    ch;
        int    d;
    }xyz;

    typedef union {
        xyz X;
        char y[100];
    }abc;

    printf("sizeof xyz = %d sizeof abc = %d\n",
                sizeof(xyz), sizeof(abc));
}

Solution for L2.Q5
L2.Q6 : Find out the error in this code

#include
#include

#define Error(str)   printf("Error : %s\n", str); exit(1);

main()
{
    int fd;
    char str[20] = "Hello! Test me";

    if ((fd = open("xx", O_CREAT | O_RDWR)) < 0)
        Error("open failed");

    if (write(fd, str, strlen(str)) < 0)
        Error("Write failed");
    if (read(fd, str, strlen(str)) < 0)
        Error("read failed");

    printf("File read : %s\n", str);
    close(fd);
}

Solution for L2.Q6
L2.Q7 : What will be the output of this program ?

#include

main()
{
    int    *a, i;

    a = (int *) malloc(10*sizeof(int));

    for (i=0; i<10; i++)
        *(a + i) = i * i;
    for (i=0; i<10; i++)
        printf("%d\t", *a++);

    free(a);
}

Solution for L2.Q7
L2.Q8 :

Write a program to calculate  number of 1's (bit) in a given
integer  number  i.e)  Number of 1's in the given  integer's
equivalent binary representation.


Solution for L2.Q8

Answers
L2.A1
Solution for L2.Q1

The output will be : 0 10 11 20 21

*s++    =>  *(s++)
*++s    =>  *(++s)
++*s    =>  ++(*s)

L2.A2
Solution for L2.Q2

Some compiler (ansi) may give warning message, but it will
compile without    errors.
The output will be : 0 1 2 3 4 and -1

L2.A3
Solution for L2.Q3

This is the  problem  about  Unions.  Unions are  similar to
structures  but it  differs  in  some  ways.  Unions  can be
assigned  only  with one  field at any time.  In this  case,
unions x and y can be  assigned  with any of the one field a
or b or c at one time.  During  initialisation  of unions it
takes  the value  (whatever  assigned  ) only for the  first
field.  So, The statement y = {100}  intialises  the union y
with field a = 100.

In this  example,  all fields of union x are  assigned  with
some  values.  But at any time only one of the  union  field
can  be  assigned.  So,  for  the  union  x the  field  c is
assigned as 21.50.

Thus, The output will be
    Union 2 : 22 22 21.50
    Union Y : 100 22 22
    ( 22 refers unpredictable results )

L2.A4
Solution for L2.Q4

The pointer x points to the same location where y is stored.
So, The changes in y reflects in x.

The output will be :
    10 11    30 31    20 21    40 41

L2.A5
Solution for L2.Q5

The  output  of  this  program  is  purely  depends  on  the
processor  architecuture.  If the sizeof  integer is 4 bytes
and the size of character is 1 byte (In some computers), the
output will be

    sizeof xyz = 20     sizeof abc = 100

The output can be generalized to some extent as follows,

    sizeof xyz = 4 * sizeof(int) + 1 * sizeof(char) +
                                            padding bytes
    sizeof abc = 100 * sizeof(char) + padding bytes

To keep the  structures/unions  byte  aligned,  some padding
bytes are added in between  the  sturcture  fields.  In this
example 3 bytes are  padded  between  ' char ch' and 'int d'
fields.  The unused bytes are called  holes.  To  understand
more about  padding bytes (holes) try varing the field types
of the structures and see the output.


L2.A6
Solution for L2.Q6

Just  try to  execute  this  file as such.  You can find out
that it will exit immediately.  Do you know why?

With  this  hint, we can trace  out the  error.  If you look
into the macro  'Error', you can easily  identify that there
are two separete statements without brases '{ ..}'.  That is
the  problem.  So, it exits  after the calling  open().  The
macro should be put inside the brases like this.

#define Error(str)     { printf("Error : %s\n", str); exit(1); }

L2.A7
Solution for L2.Q7

This program will fault (Memory  fault/segmentation  fault).
Can you predict Why?

Remove  the  statment  'free(a);'  from  the  program,  then
execute  the  program.  It will run.  It gives  the  results
correctly.

What causes 'free(a)' to generate fault?

Just trace the address  location  of pointer  variable  'a'.
The variable 'a' is incremented  inside the 'for loop'.  Out
side the 'for loop' the  variable  'a' will point to 'null'.
When the  free()  call is made, it will  free the data  area
from the  base_address  (which is passed as the  argument of
the  free  call)  upto  the  length  of the  data  allocated
previously.  In this case,  free()  tries to free the length
of 10 *sizeof(int)  from the base pointer location passed as
the argument to the free call, which is 'null' in this case.
Thus, it generates memory fault.


L2.A8
Solution for L2.Q8

#include

main(argc, argv)
int argc;
char *argv[];
{
    int count = 0, i;
    int v = atoi(argv[1]);

    for(i=0; i<8*sizeof(int); i++)
            if(v &(1<