Output of Code closed

Question

#include <stdio.h>
int main() {
    int a;
    char *x;
    x= (char *) &a;
    a=512;
    x[0]=1;
    x[1]=2;
    printf("%d\n",a);
    return 0;
}

Comments

giving output 513

even if printing x, it is giving a large output

not getting how.

@Kapil @Debashish can u plz chk it?

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Answer is correct srestha..wats the problem

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@Tauhin can u explain it more clearly?

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Srestha check it

Answer 1

Output is 513 in a little endian machine. To understand this output, let integers be stored using 16bits. In a little endian machine, when we do x[0] = 1 and x[1] = 2, then umber a is changed to 00000001 00000010 which is representation of 513 in a little endian machine.

Answer 2

#include <stdio.h>
int main() {
    int a;
    char *x;
    x= (char *) &a;// here type casting as character pointer point only to character variable
    a=512;//00000000(at x[0]) 00000001(at x[1])  as little endian
    x[0]=1;// 00000001  00000001
    x[1]=2;// 00000001  00000010
    printf("%d\n",a);  ////513
    return 0;
}

Comments

512 in little endian will be 1205 rt?

how r u getting

00000000(at x[0]) 00000001(at x[1])

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here integer is printng which wil take 2 byte  so it will take whole 2 byte to print

thats why 513

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Cannot understand, can u elaborate. what is 8 bit representation of 1 or 3?

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1 === 00000001

3 === 00000011

in little endian

if 31 ====  00000001 00000011

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but u have written in 16 bit representation.how getting 16 bit
 x[0]=1;// 00000001  00000001
    x[1]=2;// 00000001  00000010

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8- 8 bit pointed by character pointer which is divide from integer to character while type casting

these is my concept

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can u give me some link? still not getting.

do u added bitwise x[1],x[2]?

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what is ur doubt i can't understand?

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http://cs-fundamentals.com/tech-interview/c/c-program-to-check-little-and-big-endian-architecture.php

visit this

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visited the link

There are so many problems in this question, first tell me

a=512

Now x[0]=1

x[1]=2

So, x[]=21

Now, we are printing a

a must be same , So, why not it printing 512?

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Am I wrong somehow?

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char *x;
    x= (char *) &a;

here take a look is an integer variable so now it will be pointed by character pointer thats why type casting is done as not character is of 1 byte .

now,   

a= 512 = 00000001    00000000

so as per little endian the MSB will filled in LSB side and vice versa

so a= 00000000  00000001

now X[0]=1 , means as X  is a character array so for this MSB will be X[0]

therefore new a= 00000001   00000001

now X[1]=2 , means as X  is a character array so for this right from  MSB will be X[1]

therefore new a = 00000001 00000011

now as a execution of %d come which take a as a whole 2 byte so for this it will take MSB first and first and

as u know in integer it will be 16 bit there for it conscider whole 16 bit 

print a=(00000001  0000001{x[1]}1{x[0]}) = 513

hope it will be helpful for u