Geeksquiz | Array

Array

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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 a and a size of integer//is a arr++;            printf ("%u", arr); return 0; }

A	2002
	2004
C	By 2020
	Lvalue Required

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Question 1 Explanation:array name in C was 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 in a and a size of integer is the + 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 is used with array name

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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-ofOperator before the array name gives the address of the 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

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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

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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) ". The and this value are typecasted to int *. So ptr points the memory just after 6 are stored. PTR is typecasted to "int *" and value of * (PTR-1) is printed. Since ptr points memory after 6, ptr–1 points to 6.