CSE Pointers Quiz-1

Parameters are always passed by value in C. Therefore, in the above code, value of y is not modified using the function fun(). So how do we modify the value of a local variable of a function inside another function. Pointer is the solution to such problems. Using pointers, we can modify a local variable of a function inside another function. See the next question. Note that everything is passed by value in C. We only get the effect of pass by reference using pointers.

The function fun() expects a pointer ptr to an integer (or an address of an integer). It modifies the value at the address ptr. The dereference operator * is used to access the value at an address. In the statement ‘*ptr = 30’, value at address ptr is changed to 30. The address operator & is used to get the address of a variable of any data type. In the function call statement ‘fun(&y)’, address of y is passed so that y can be modified using its address.

Each letter in string[] array is stored in seperate address. The starting address in string[] array is stored a pointer variable ptr which is then incremented by 6. Thus "learn " is neglected and "C from 2braces.com" is displayed.

Answer : Option (D)

size of void is 1 and the size of void pointer is 4

int *ptr; /* Note: the use of * here is not for dereferencing,

it is for data type int */

;

ptr = &x; /* ptr now points to x (or ptr is equal to address of x) */

= 0; /* set value ate ptr to 0 or set x to zero */

f(" x = %d\n", x); /* prints x = 0 */

printf(" *ptr = %d\n", *ptr); /* prints *ptr = 0 */

+= 5;          /* increment the value at ptr by 5 */

printf(" x = %d\n", x); /* prints x = 5 */

printf(" *ptr = %d\n", *ptr); /* prints *ptr = 5 */

(*ptr)++;          /* increment the value at ptr by 1 */

printf(" x = %d\n", x); /* prints x = 6 */

printf(" *ptr = %d\n", *ptr); /* prints *ptr = 6 */

Here the pointer variable stores the address of first value in struct node num = {3, 5, 6}; which is incremented by 2 and then the value 6 is finally displayed.

Answer-4

When we add a value x to a pointer p, the value of the resultant expression is p + x*sizeof(*p) where sizeof(*p) means size of data type pointed by p. That is why ptr2 is incremented to point to arr[3] in the above code. Same rule applies for subtraction. Note that only integral values can be added or subtracted from a pointer. We can also subtract or compare two pointers of same type.

Array name gives the address of first element in array. So when we do '*ptr1 = arr;', ptr1 starts holding the address of element 10. 'arr + 5' gives the address of 6th element as arithmetic is done using pointers. So 'ptr2-ptr1' gives 5. When we do '(char *)ptr2', ptr2 is type-casted to char pointer and size of character is one byte, pointer arithmetic happens considering character pointers. So we get 5*sizeof(int)/sizeof(char) as a difference of two pointers.