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

Why are only 7 bits being stored in a byte?

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Hi @ ManojK     can you explain what would be the output of this program if we remove the little-endian processor.   condition  would that be changed ? or be same as it is .

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Just run the program .

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@shekhar obviously. It will then print the most significant byte. But big-endian machines are rare.

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@arjun sir and @manojK I ran it again without the condition and o/p is still 64

but my confusion is in there 2 lines

  ptr=(char *)&a;
    printf("%d",*ptr);

a has a value 320 . in first line we are type casting  the address of the value  a , just for example let say the address of a is 100 (this is what i understood from this how do i know what is the address of a and what would be the value of it after type casting into char) now first line typecasting that short value in char and storing into ptr .now how am i getting 64 ?

the compiler did not say anything why it gave 64 to me.

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 let say the address of a is 100 (this is what i understood from this how do i know what is the address of a and what would be the value of it after type casting into char)

It'll be still 100 - address has same value for ant type. Type casting here is for ensuring type safety. 

now how am i getting 64

Due to *ptr as type of ptr is char *. Try *(int*)ptr. 

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@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?

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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.