Chapter - Pointers
Ans 11.4 Pointer is initialized by keyword NULL or 0 in
the following fashion
int * p = NULL , * m = 0 ;
Ans 11.5 Following is the effect of the following
statements
i.
int a, *b = &a; //Error Pointer being initialized with the
address of an uninitialized variable
ii.
int p ,
*p ; //Error Multiple
Declaration of p
iii.
char *s;
iv.
a = (float * ) &x);
v.
double ( *f
)( ) ; //It declares the pointer to
a function
Ans 11. 6 int M
, n ;
int *P1
, *P2 ;
The following are the errors in
the expression given below:
i)
P1 = &m ; //Error Pointer cannot be
initialized with an uninitialized variable
ii)
P2 = n ; // Error Pointer can only be
initialized with the address of the variable
iii)
P2 = &*&m; // Correct as the first
dereferencing operator Is over powered by the second & operator
iv)
M = P2 –
P1 //Correct Pointer Subtraction Is allowed
v)
P1 = &P2 //Error A simple pointer cannot be
initialized with memory address of an another pointer, only a pointer to a
pointer can be initialized in this way
Ans11.7 Difference between the notations
*m[5] and (*m)[3]. Since * has lower precedence than [] , *m[5] declares m as an array of 5 pointers while (*m)[3]
declares m as a pointer to an array of five elements.
Ans 11.9 int x = 10
, y
= 10 ;
int
*p1 = &x , *p2 = &y;
The
following Expression generate the following output:
i)
(*p1)++
Output – 10
ii)
- - (*p2)
Output - 9
iii)
*p1 + (*p2)--
Output - 20
iv)
++(*p2)
- *p1 Output - 1
Ans 11.12 Following
Output is generated by the following
code :
int m =
100 ;
int *p1
= &m;
int
**p2 = &p1;
printf(“%d”
, **p2);
Output
:
100
Ans 11. 13 Following
is the error in the given code
int
**p1 , * p2 ;
p2
= &p1 // It should be p1 =&p2
Explanation: p1 is a pointer to a pointer so it should
always point a pointer
Ans11.14 Difference
between the two expression name + 10; and *(name + 10 ) is that the first
expression refers to all the characters after the 10th position in
the array name or rather simply we can
say that it will the array name from the 10th character whereas the
second expression refers specifically to the 10th character of the
array of name .
Ans 11.15 Following
is the output of the code given below
int
m[2] ;
*(m+1)
*m
= *(m + 1);
printf(“
%d “ , m[0] );
Output:
100
Ans 11.16 int m[2];
int *p = m ;
m[0] = 100;
m[1] = 200;
printf(“%d %d “ , ++*p , *p );
Output:
101
100
Ans 11.17 Output of
the following code :
int f(char *p );
main()
{ char str[] = “ANSI”;
printf(“%d”
, f(str) );
}
int
f(char *p )
{ char *q = p ;
while
(*++p ) ;
return
(p – q ); }
Output :
4
Ans 11.18 Following
declarations are not the same:
1 .
void search ( int *m[] , int x) { }
In this
the parameter in the definition is a integer pointer array.
2. void
search ( int **m ,int x ) {}
In this
the parameter in the definition is a integer pointer to a pointer.
Ans 11.19 Following
declaration is are same
char
*s;
char
s[5] ;
As name
of the array is a pointer to the first element of the array it’s as same as
declaring a pointers which when incremented can be used to store a list of
values same as an array.
Ans 11.20 The
following is the correct way to declare a pointer to a function.
int
(*p)(void);
