C programming

Question

#include <stdio.h>

int main()
{
    int i =258;
    int *iptr = &i;
    printf("%d%d",*((char*)iptr),*((char*)iptr+1));

    return 0;
}

(a)2,1 (b)2,5

(c)1, 2  (d)5,2

I am getting  (a) as output

sh-4.2$ gcc -o main *.c                                                                                                                                

sh-4.2$ main                                                                                                                                           

21

       

Can anybody explain that how we got that?

Thanks :)                                                                                                                                     

           

Comments

yes output will be 21...

default we take little endian machine..., in little endian machine higher bytes are saved in increasing memory location..

for eg(take int as 2 bytes)..

int a=258, in binary 2's complement its value will be,

0000000100000010, now according to little endian higher bytes(in bold) will be stored in higher memory location, and the corresponding number will be saved as 0000001000000001 

now its easy you can do it, just apply pointers rules you will get the answer as 21..

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Thanks :)

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@joshi_nitish 

Please explain the last paragraph of your answer and make it full understandable.