This document discusses pointers to pointers, arrays to pointers, and pointers to arrays. It explains that a pointer to a pointer contains the address of another pointer, requiring double indirection to access the target value. An array name is a constant pointer to the first element. An array can be accessed using pointers by making a pointer point to the base address of the array. Pointers are useful for handling ragged arrays of strings of varying lengths more efficiently than fixed-length character arrays.

1. NAME:- DARJI MIT BIPINBHAI
TOPICK : - POINTER TO POINTER
POINTER TO ARRAY
ARRAY TO POINTER

2. POINTER TO POINTER
 A pointer to a pointer is a form of multiple
indirection, or a chain of pointers. Normally, a pointer
contains the address of a variable.
 When we define a pointer to a pointer, the first pointer
contains the address of the second pointer, which
points to the location that contains the actual value as
shown below.
 A pointer to a pointer is a form of multiple indirection,
or a chain of pointers. Normally, a pointer contains the
address of a variable. When we define a pointer to a
pointer, the first pointer contains the address of the
second pointer, which points to the location that
contains the actual value as shown below.

3. .
−
I ptr1
ptr2

4.  A variable that is a pointer to a pointer must be
declared as such. This is done by placing an additional
asterisk in front of its name. For example, the following
declaration declares a pointer to a pointer of type int −
int **var
When a target value is indirectly pointed to by a pointer to a
pointer, accessing that value requires that the asterisk
operator be applied twice, as is show below in the example-;

5. #include <stdio.h>
Int main (){
int var ;
int*ptr;
int**pptr;
var=3003
/*take the address of var */
ptr = &var;
/*take the address of ptr using address of operator&*/
pptr =&ptr;
/*take the value using pptr*/
printf(“value of var %dn”,var);
print f (“value available at *ptr =%dn,*ptr);
print f(“value available at **pptr =%dn”,**pptr);
return 𝜃;
}
When the above code is compiled and executes , it produces the
following result

6.  When an array is declared, the compiler allocates a base
address and sufficient amount of storage is the location
of the first element(index 0) of the array. The compiler also
defines the array name as a constant pointer to the first
element. Suppose we declare an array x as follows:
int X[5]={1,2,3,4,5};
 Suppose the base address of x is 1000 and assuming
that each integer requires two bytes, the five elements
will be stored as follows:
Elements X[0] X[1] X[2] X[3]
X[4]
value
Address
Base address
the name X is defined as a constant pointer pointing to the first element,
X[0] and therefore the value of X is 1000, the location where X[0] is stored. This
is,
X= &x [0] =1000
1 1000 2 1002 3 1004 4 1008

7.  if we declare p as an integer pointer, then we can
make the pointer p to point to the array x by the
following assignment:
p=x;
This is equivalent to
p=&x[0];
 Now, we can access every value of X using p++ to
move from one element to another. The relationship
between p and X is shown as :
p=&x[0](=1000)
p+1=&x[1](=1002)
p+2=&x[2](=1004)
p+3=&x[3](=1006)
p+4=&x[4](=1008)

8.  You may notice that the address of an
element is calculated using its index and the
scale factor of the data type. For instance,
address of x[3] = base address+(3x scale factor of int)
=1000+(3×2) 1006
 When handing array, instead of using array
indexing, we can pointers to access array elements.
Note that *(p+3) gives the value of x[3]. The pointer
accessing method is much faster than array indexing.

9.  One important use of pointer is in handling of a
table of strings. Consider the following array of
strings:
char name [3] [25];
 This says that the name is a table containing three
names, each with a maximum length of 25
characters (including null character). The total
storage requirements for the name table are 75
bytes.
 We know that rarely the individual strings will be of
equal lengths. Therefore, instead of making each row
a fixed number of characters, we can make it a
pointer to a string of varying length.

10. For example,
char *name[3] = {
“New Zeland”,
Australia”,
“India”
};
declares name to be an array of three pointers to
characters, each pointer pointing to a particular name as:
name[0] New zeland
name[1] Australia
name[2] India
This declaration allocates only 28 bytes, sufficient to hold
all the characters as shown

11. The following statement would print out all the three
names;
for(i=0; i<=2; i++)
printf(“%sn “, name[i]);
 To access the jth character in the ith name, we may
write as
*(name[i]+j)
 The character array with the rows of varying length
are called ragged array’ and are better handled by
pointers.
 Remember the difference between the notations *p[3]
and (*p)[3].Since *has a lower precedence than [],*p[3]
declares p as an array of 3 pointers while (*p)[3]
declares p as a pointer to an array of three elements.
N e w Z e a l a n d 0
A u s t r a l i a 0
I n d i a 0