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Question

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

Answer 1

output will be machine dependent .

on little endian machine output will be 513

Comments

@srestha yes according to little endian 513.If nothing is given we follow the little endian but output will be different in case of big endian.

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I have doubt:- Why this code is not showing segmentation fault.

#include <stdio.h>
int main()
{
    printf("%d\n%d\n", sizeof(int), sizeof(char));
    int a = 512;
    char *p;
    p = (char *)&a;
    p[0] = 1;
    p[1] = 0;
    p[2] = 0;
    p[3] = 0;
    printf("%d",a);
    p[4] = 110;
    p[5] = 10;
    printf("\n%d",a);
    return 0;
}

Since, int is 4 byte and char is 1 byte, according to me it should show segmentation fault at p[4] = 110 and also its below line since we are accessing the location which is not allowed to access.

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C has no bound checking therefore if you have a[2], you can access a[20] or a[76] etc....

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i am not sure on these points :-

 

in C, there is nothing called " the location which is not allowed to access. "

it will give segmentation fault, when you write in read-only location, but if you want read in read-only location, you can do that.