POINTERS

1.a) POINTER EXPRESSIONS Pointer variables can be used in expression If p1 and p2 are properly declared and initialized pointers then following are valid y = *p1 * *p1; same as (*p1)*(*p2) sum=sum + *p1; z = 5* - *p2/ *p1; same as (5 * (-(*p2)))/(*p1) p1++ -p2 sum += *p2

Pointers can also be compared using relational operators. p1 > p2, p1 == p2, p1 != p2 Pointers cannot be divided or multiplied or added p1/p2 or p1*p2 or p1/3 or p1+p2 Add integer to or subtract integers from pointers are allowed Subtraction of one pointer from another of same type is allowed

Important note *ptr++ is equivalent to *(ptr++) as ++ has greater operator precedence expression will increase the value of ptr so that it points to the next element. (*ptr)++ will increment the value of the variable whose address is stored in ptr.

1.b) POINTER INCREMENT P1++ or p1=p1+1 will cause the pointer to point to the next value of its type. e.g if p1 value is 2800 then after p1++ it will be *p1 will give value stored at next location The value by which it is increased is called as scale factor Sizeof(variable) – operator returns size of bytes occupied by that variable

#include int main() { int num, *pnum; pnum=# *pnum=10; printf("*pnum= %d", *pnum); printf(" num = %d", num); *pnum= *pnum + 1; //increments the value of num printf("\n after increment *pnum=%d", *pnum); printf("\n after increment num=%d", num); getch(); return 0; } Find out output.

1.d) POINTERS AND CHARACTER STRINGS String without pointer char str[5]=“good” String with pointer char *str=“good”; pointer str now points to the first character of the string or char *string1; string1= “good”; string1 is not a string it is pointer which also will work as a name for the string To print string string1 we can use printf or puts (). * not required

/* STRING HANDLING USING POINTERS */ void main() { char *name; int length; char *cptr; name = "DELHI"; cptr=name; printf ("%s\n", name); while(*cptr != '\0') { printf("%c is stored at address %u\n", *cptr, cptr); cptr++; } length = cptr - name; printf("\nLength of the string = %d\n", length); getch(); } Output:- DELHI D is stored at address 170 E is stored at address 171 L is stored at address 172 H is stored at address 173 I is stored at address 174 Length of the string = 5

#include int main() { // Declaring/Initializing three characters pointers char *ptr1 = "Himanshu"; char *ptr2 = "Arora"; char *ptr3 = "TheGeekStuff"; //Declaring an array of 3 char pointers char* arr[3]; // Initializing the array with values arr[0] = ptr1; arr[1] = ptr2; arr[2] = ptr3; //Printing the values stored in array printf("\n [%s]\n", arr[0]); printf("\n [%s]\n", arr[1]); printf("\n [%s]\n", arr[2]); getchar(); return 0; }

1.c) POINTERS AND 1D- ARRAYS Accessing array elements by pointers is always faster than accessing them by subscripts. e.g int x[5] = { 1,2,3,4,5}; Int *p; We can make pointer p point to an array x by: p = x or p = &x[0] 1D, x[i] can be represented by *(x+i) or *(p+i)

Array notation is a form of pointer notation. The name of the array is the starting address of the array in memory. It is also known as a base address. In other words, base address is the address of the first element in the array or the address of arr[0]. Int *ptr; Ptr = &arr[0]; Here, ptr is made to point to the first element of the array. Ptr++; // will point to the second element of the array

Impt : Num[i], *(num+i), *(i+num), i[num] -- all refer to same element int main() { int num[ ] = { 24, 34, 12, 44, 56, 17 } ; int i ; for ( i = 0 ; i <= 5 ; i++ ) { printf ( "\naddress = %u ", &num[i] ) ; printf ( "element = %d %d ", num[i], *( num + i ) ) ; printf ( "%d %d", *( i + num ), i[num] ) ; } getch(); return 0; } This output is on 64-bit pc due to which int is taking 4 byte

int main() { int num[ ] = { 24, 34, 12, 44, 56, 17 } ; int i, *j ; j = &num[0] ; /* assign address of zeroth element */ for ( i = 0 ; i <= 5 ; i++ ) { printf ( "\naddress = %u ", j ) ; printf ( "element = %d", *j ) ; j++ ; /* increment pointer to point to next location */ } getch(); return 0; } This output is on 64-bit pc due to which int is taking 4 byte

E.g main() { int b[]={ 10,20,30,40,50}; int i, *k; k=b; for(i=0;i<=4;i++) { Printf(“\n%d”, *k); k++; } what will be output ??????

1.d) POINTERS AND 2D ARRAY Int b[3][3] = { {1,3,5}, {7,9,11}, {13,15,17}} b is a pointer to b[0][0] b[0] is also a pointer to b[0][0] b[1] is a pointer to b[1][0] b[2] is a pointer to b[2][0] *b is a pointer to b[0] (special case) *b[1] is the value of b[1][0] which is 7 *b[2]+1 is the value of b[2][0]+1 (which is 14)

Int b[10]; we can write b[0] as *(b+0) or *b Int a[0][0] can also be written as *(a[0]+0) or *(*a+0) or **a S[2][1] *(s[2]+1) *(*(s+2)+1) all refer to same element Fig. memory map of a 2D array

OUTPUT

Exercise : find out the output for the program and explain working of program

1.e) POINTERS AND 3D ARRAY Arr[2][3][1] is written as *(*(*arr+2)+3)+1)

1.f) ARRAY OF POINTERS Like array of ints or an array of floats, we have array of pointers Array of pointers is a collection of addresses. The addresses present in the array of pointers can be : a) addresses of isolated variables b) addresses of array element c) or any other addresses

Example 1 : Array of pointer

Example 2 : Array of pointers // run this code on software Output ????????? Find by yourself

Questions) using pointers WAP to find largest number in an array using pointer WAP to find smallest number in an array using pointer WAP to exchange the values of largest with the smallest number of an array WAP to find the sum of all elements of an array WAP to read and display the values of 3*3 matrix WAP to read and display the values of 3*3*3 matrix

1.f) POINTER AND STRUCTURE Pointer can also point to a structure. Such pointers are called as ‘structure pointers’ Structure variables can be accessed by ‘->’ operator or dot (.) operator Pointer_var->structure_var (*pointer_var).structure_var

Example : structure pointer