ISRO2016-25

Question

What will be output of the following program? Assume that you are running this program in little-endian processor.

#include<stdio.h>
int main()
{
    short a=320;
    char *ptr;
    ptr=(char *)&a;
    printf("%d",*ptr);
    return 0;
}

• A. 1
• B. 320
• C. 64
• D. Compilation error

Answer 1

 short a=320;

how a will be represented in binary

   HB                                                                                LB

00000001

01000000

Now according to Little-endian .The least significant byte (LSB) value , is at the lowest address. The other bytes follow in increasing order of significance.

Now

 ptr=(char *)&a;

a is type casted to char so according to Little endianness .The least significant byte (LSB) value , is at the lowest address.

So ptr will be stored like this

   LA

01000000

   So when the printf is executed it will print value 64.

So C is correct.

Comments

@Arjun Sir why here little endian store least significant byte , not most significant byte?

In little endian there is an address ordering . So how starting address will be same for little endian too?

---

srestha let value   ABCDE.....we want to store...in memory   in this MSB = A , LSB = E

BIG- ENDIAN let addressres be 00 ,01,02.....(though it is in binary format in computer but for clearer view assume adrress in decimal here)

00	A
01	B
02	C
03	D
04	E

LITTLE ENDIAN

00	E
01	D
02	C
03	B
04	A

address is only in increasing order...but the values that we store at locations need order according to mechanism  we follow..

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@habedo007 because "short is of size 2B" while it's type-casted to "char which is of 1B".

Answer 2

answer is c)64

Comments

explain plz.