C questions on Pointers, Storage class specifier

1. Mohammed Sikander
Technical Lead
CranesVarsity (www.cranesvarsity.com)
Mohammed.sikander@cranessoftware.com

2. int countNodes( Node *root)
{
if(root == NULL)
return 0;
int leftNodes = countNodes(root->left);
int rightNodes = countNodes(root->right);
return leftNodes + rightNodes + 1; //+1 for current node
}

3. int countLeaf( Node *root)
{
if(root == NULL)
return 0;
if(root->left == NULL && root->right == NULL)
return 1;
int leftLeafNodes = countLeaf(root->left);
int rightLeafNodes = countLeaf(root->right);
return leftLeafNodes + rightLeafNodes;
}

4. int calculateHeight( Node *root)
{
if(root == NULL)
return 0;
int leftHeight = calculateHeight(root->left);
int rightHeight = calculateHeight (root->right);
return max(leftHeight , rightHeight) + 1;
}

5. FILE1.C
int x = 5;
void print( )
int main( )
{
print( );
printf(“Main x = %d n”,x);
}
FILE2.C
extern int x;
void print( )
{
printf(“Print : x = %d n”,x);
}

6. FILE1.C
extern int x;
void print( );
int main( )
{
printf(“Main x = %d n”,x);
print( );
}
FILE2.C
int x = 5;
void print( )
{
printf(“Print : x = %d n”,x);
}

7. Learning : We require one definition and a declaration in
other file(s). It does not matter which file contains the
definition.

8. FILE1.C
int x = 8;
void print( );
int main( )
{
printf(“Main x = %d n”,x);
print( );
}
FILE2.C
int x = 5;
void print( )
{
printf(“Print : x = %d n”,x);
}

9.  Multiple Definition Error.
 An external declaration for an object is a
definition if it has an initializer (Refer K&R
A10.2 ).
 As both have initialization, both are
definition.
 We will get Linker Error (not a compile-time
error)

10. FILE1.C
static int x;
void print( );
int main( )
{
printf(“Main x = %d n”,x);
print( );
}
FILE2.C
int x = 5;
void print( )
{
printf(“Print : x = %d n”,x);
}

11.  X in file1 has internal linkage and is not visible
during linking of multiple files.
 No error of multiple definition.
 File1 will access the variable x defined in file1,
as it is not initialized, the default value is 0.
 File2 will access the variable x defined in file2,
whose value is 5

12. FILE1.C
static int x = 8;
void print( );
int main( )
{
extern int x;
printf(“Main x = %d n”,x);
print( );
}
Output is 8.
Extern says that the variable
is defined externally, within
the same file / other file. As
an external definition is
found, it refers to it.

13. FILE1.C
int x ;
void print( );
int main( )
{
printf(“Main x = %d n”,x);
print( );
}
FILE2.C
int x = 5;
void print( )
{
printf(“Print : x = %d n”,x);
}

14.  An external object declaration that does not have an
initializer, and does not contain the extern specifier, is a
tentative definition
 If a definition for an object appears , any tentative definitions
are treated merely as redundant declarations. If no definition
for the object appears, all its tentative definitions become a
single definition with initializer 0.
 The declaration of x in file1 is tentative definition;
 As the definition for x is found in file2, x in file1 is treated as
declaration only.

15. FILE1.C
int x ;
void print( );
int main( )
{
printf(“Main x = %d n”,x);
print( );
printf(“Main x = %d n”,x);
}
FILE2.C
int x;
void print( )
{
x = 5;
printf(“Print : x = %d n”,x);
}

17. FILE1.C
int arr[5] = {2,5,7};
void print( );
int main( )
{
print( );
}
FILE2.C
extern int arr[ ];
void print( )
{
printf(“ %d “ , arr[0] );
}

18. FILE1.C
int arr[5] = {2,5,7};
void print( );
int main( )
{
print( );
}
FILE2.C
extern int *arr;
void print( )
{
printf(“ %d “ , arr[0] );
}

19. FILE1.C
int arr[5];
void print( );
int main( )
{
print( );
}
FILE2.C
extern int arr[ ];
void print( )
{
printf(“ %d “ , sizeof(arr) );
}

20. FILE1.C
int arr[5] = {2 , 5 , 8};
void print(int arr[]);
int main( )
{
print( arr );
}
FILE2.C
void print(int arr[] )
{
printf(“ %d “ , sizeof(arr) );
}

21. FILE1.C
int arr[5] = {2 , 5 , 8};
void print(int arr[]);
int main( )
{
print( arr );
}
FILE2.C
void print(int *arr )
{
printf(“ %d “ , arr[0] );
}

22. FILE1.C
char arr[25] = “RADHA”;
void print( );
int main( )
{
print( );
}
FILE2.C
extern char *arr;
void print( )
{
printf(“ %s “ , arr);
}

23.  Draw an Expression Tree for the given
expression and traverse in inorder,preorder
and postorder.
 A + B * C
 A * (B + C)

24. int *pi;
char *pc;
short int *psi;
printf(“ %d n”,sizeof(pi));
printf(“ %d n”,sizeof(pc));
printf(“ %d n”,sizeof(psi));
int *pi;
char *pc;
short int *psi;
printf(“ %d n”,sizeof(*pi));
printf(“ %d n”,sizeof(*pc));
printf(“ %d n”,sizeof(*psi));

25. int *pi = (int *)2000;
char *pc = (char *)2000;
pi++;
pc++;
printf(“ %u %u “ , pi , pc);

26. int x = 0x1234;
char c = x;
printf(“ %x “ , c);

27. const int x = 10;
1.int * const ptr1 = &x;
2.int const * ptr2 = &x;
3.const int * ptr3 = &x;

28. void update(int *p)
{
*p = *p + 5;
}
int main( )
{
int x = 10;
int *ptr = &x;
update(ptr);
printf(“x = %d “ , x);
}

29. void update(int *p)
{
p = p + 1;
}
int main( )
{
int arr[ ] = {10 , 12 , 25 , 45};
int *ptr = arr;
update(ptr);
printf(“*ptr = %d “ , *ptr);
}

30. void update(int *p)
{
*p = *p + 1;
}
int main( )
{
int arr[ ] = {10 , 12 , 25 , 45};
int *ptr = arr;
update(ptr);
printf(“*ptr = %d “ , *ptr);
}

31. void update(char *str)
{
str = “Devendra”;
}
int main( )
{
char *name = “Nimisha”;
update(name);
printf(“ %s n” , name);
}

32. void update(char *str)
{
str[0] = ‘H’;
}
int main( )
{
char name[ ] = “Nimisha”;
char *ptr = name;
update(ptr);
printf(“ %s n” , name);
}

33. void update(char *str)
{
*++str = ‘a’;
}
int main( )
{
char name[ ] = “Nimisha”;
char *ptr = name;
update(ptr);
printf(“ %s n” , name);
}

34. int main( )
{
char names[]
[10]={“OBAMA”,”PUTIN”,”MODI”,”CAMEROON”};
printf(“ %__ “ , *(names + 2) + 3);
printf(“ %__ “ , **(names + 2) + 3);
}

35. int main( )
{
char names[][10]={“OBAMA”,”PUTIN”,”MODI”,”CAMEROON”};
1. printf(“ %s “ , *(++names));
2. printf(“ %s “ , ++*(names));
3. printf(“ %c “ , ++**(names));
}

36. int main( )
{
char *names[10]={“OBAMA”,”PUTIN”,”MODI”,”CAMEROON”};
1. printf(“ %s “ , *(++names));
2. printf(“ %s “ , ++*(names));
3. printf(“ %c “ , ++**(names));
}

37. struct Student
{
int id;
char name[20];
};
int main( )
{
struct Student s1 = {1 , “Ayushi”};
struct Student s2 = {2 , “Ayushi”};
if(s1.name == s2.name)
printf(“Equal”);
else
printf(“Not Equal”);
}

38. char *names[ ] = {“Nimisha”,”Devender”,”Vikram”,”Balwant”};
printf(“ %d “ , sizeof(names));
printf(“ %d “ , sizeof(names[0]));
char names[ ][10] = {“Nimisha”,”Devender”,”Vikram”,”Balwant”};
printf(“ %d “ , sizeof(names));
printf(“ %d “ , sizeof(names[0]));

39. void display( char *names[ ])
{
printf(“Display : %d n “ , sizeof(names));
}
int main( )
{
char *names[ ] = {“Nimisha”,”Devender”,”Vikram”,”Balwant”};
printf(“Main : %d n“ , sizeof(names));
display( names );
}

40. static int x = 5;
int main( )
{
int x;
printf(“x = %d n”,x);
}

41. static int x = 5;
int main( )
{
extern int x;
printf(“x = %d n”,x);
}

42. int a = 5;
int b;
static int c = 7;
static int d;
int main( )
{
}
• Mention the memory segments of
variables after compiling (.o file) and
after linking (a.out)
• What is the difference between a
and c.
• Use nm utility to view the memory
segments
$gcc –c file.c
$nm file.o
$gcc file.c
$nm ./a.out

43. void func( );
int x;
int main( )
{
x = 5;
func( );
printf(“x = %d “ , x);
}
int x;
void func( )
{
x = 10;
}

44. [sikander@localhost ~]$ gcc -c f1.c
[sikander@localhost ~]$ gcc -c f2.c
[sikander@localhost ~]$ nm f1.o
U func
00000000 T main
U printf
00000004 C x
[sikander@localhost ~]$ nm f2.o
00000000 T func
00000004 C x
[sikander@localhost ~]$ gcc f1.o f2.o
[sikander@localhost ~]$ nm a.out
080495ac B x

45. void func( );
int x = 5;
int main( )
{
func( );
printf(“x = %d “ , x);
}
int x = 10;
void func( )
{
x++;
}

46.  [sikander@localhost ~]$ gcc -c f1.c
 [sikander@localhost ~]$ gcc -c f2.c
 [sikander@localhost ~]$ nm f1.o
 00000000 D x
 [sikander@localhost ~]$ nm f2.o
 00000000 D x
 [sikander@localhost ~]$ gcc f1.o f2.o
 f2.o(.data+0x0): multiple definition of `x'
 f1.o(.data+0x0): first defined here
 collect2: ld returned 1 exit status

47. C C
B
C D
D
D D
Multiple Definition
d d
No Conflict, two different variables
b b
No Conflict, two different variables
d b
No Conflict, two different variables
b d
No Conflict, two different variables
b D
No Conflict, two different variables

48. int x ; //C
int main( )
{
printf(“Main &x= %p” , &x);
printf(“x = %d n”, x);
func( );
}
int x ; //C
void func( )
{
printf(“Func &x= %p” , &x);
printf(“x = %d n”, x);
}
int x ; //C
int main( )
{
printf(“Main &x= %p” , &x);
printf(“x = %d n”, x);
func( );
}
int x = 5 ; //D
void func( )
{
printf(“Func &x= %p” , &x);
printf(“x = %d n”, x);
}

49. void func( );
int x = 5; //D
int main( )
{
printf(“Main &x= %p” , &x);
printf(“x = %d n”, x);
func( );
}
int x = 10; //D
void func( )
{
printf(“Func &x= %p” , &x);
printf(“x = %d n”, x);
}
void func( );
static int x = 5; //d
int main( )
{
printf(“Main &x= %p” , &x);
printf(“x = %d n”, x);
func( );
}
int x = 10; //D
void func( )
{
printf(“Func &x= %p” , &x);
printf(“x = %d n”, x);
}

50. const int x = 10; //Read Only
int main( )
{
const int y = 5; //Stack
}

51. const int x = 10; //Read Only
int main( )
{
const int y = 5; //Stack
printf(“Enter the value for local const variable : “);
scanf(“ %d”,&y);
printf(“Local Const = %d n” , y);
printf(“Enter the value for global const variable : “);
scanf(“ %d”,&x);
printf(“Global Const = %d n”,x);
}
[sikander@localhost ~]$ ./a.out
Enter the value for local const variable : 89
Local Const = 89
Enter the value for global const variable : 6
Segmentation fault

52.  [sikander@localhost ~]$ nm f1.o
 00000000 t display
 0000000a T main
 00000005 T print
static void display()
{
}
void print()
{
}
int main( )
{
display( );
print( );
}