Geeksquiz | Array

Array

-----------

Arrays

Question 1 WRONG

Predict the output of below program:

#include <stdio.h>   int main() {      int arr[5];            // Assume that base address of arr is 2000 and size of integer          // is 32 bit      arr++;      printf ( "%u" , arr);            return 0; }

A	2002
	2004
C	2020
	lvalue required

Discuss it


Question 1 Explanation:  Array name in C is implemented by a constant pointer. It is not possible to apply increment and decrement on constant types.

Question 2 WRONG

Predict the output of below program:

#include <stdio.h>   int main() {      int arr[5];      // Assume base address of arr is 2000 and size of integer is 32 bit      printf ( "%u %u" , arr + 1, &arr + 1);        return 0; }

	2004 2020
B	2004 2004
	2004 Garbage value
D	The program fails to compile because Address-of operator cannot be used with array name

Discuss it


Question 2 Explanation:  Name of array in C gives the address(except in sizeof operator) of the first element. Adding 1 to this address gives the  address plus the sizeof typethe array has. Applying the  Address-of operator before the array name gives the address of the whole array. Adding 1 to this address gives the address plus the sizeof whole array.

Question 3 CORRECT

What is output?

# include <stdio.h>   void print( int arr[]) {     int n = sizeof (arr)/ sizeof (arr[0]);     int i;     for (i = 0; i < n; i++)        printf ( "%d " , arr[i]); }   int main() {     int arr[] = {1, 2, 3, 4, 5, 6, 7, 8};     print(arr);     return 0; }

A	1, 2, 3, 4, 5, 6, 7, 8
B	Compiler Error
	1
D	Run Time Error

Discuss it


Question 3 Explanation:  See  http://www.geeksforgeeks.org/using-sizof-operator-with-array-paratmeters/ for explanation.

Question 4 CORRECT

Output of following program?

#include<stdio.h>     int main() {    int a[] = {1, 2, 3, 4, 5, 6};    int *ptr = ( int *)(&a+1);    printf ( "%d " , *(ptr-1) );    return 0; }

A	1
B	2
	6
D	Runtime Error

Discuss it


Question 4 Explanation:  &a is address of the whole array a[]. If we add 1 to &a, we get “base address of a[] + sizeof(a)”. And this value is typecasted to int *. So ptr points the memory just after 6 is stored. ptr is typecasted to "int *" and value of *(ptr-1) is printed. Since ptr points memory after 6, ptr – 1 points to 6.