Answer / gaurav bhandarkar

Author : Gaurav M. Bhandarkar
Yes u can modify constants...

Theory:
const int z = 420; // z is constant
&z (address of constant)
z (value of constant)

imp:
*(int*)((char*)&(*((char*)&z+1))-1) is
a unity/identity pointer operation resulting in z
That is:-
printf("%d | %d",*(int*)((char*)&(*((char*)&z+1))-1),z);
OUTPUT: 420 | 420

code:

const int z = 420;

printf("%d | %d\n",*(int*)((char*)&(*((char*)&z+1))-1),z);
//o-p 420 | 420


*((char *)&z+1) = 21; //corrupting the constant


printf("%d | %d",*(int*)((char*)&(*((char*)&z+1))-1),z);
//o-p 5540 | 420

___
The 2 similar printf's(check they are same)
o/p different values for same "z"
which is a constant!
thus the constant was corrupted!

Answer / saneesh a t

Actually the keyword const is not for the programmer who work with the normal
general purpose computer. By declaring a variable as const, the compailar shows
warning/error to the modification to the variable. At runtime you can modify the memory location
using any technique, or an external hacking program can change the value of the variable.
Suppose your program is compiled to run with a microcontroller with flash or EEPROM or such
memory. Now your variable with const qualifier will be stored in the FLASH of EEPROM memory,
which cann't be modified, and the technique is used to save the RAM space which is too small in
size for a micro controller. In this case too, a FLASH/EEPROM write can chage the value of the
const varriable.

Answer / vignesh1988i

const. variable cannot be modified........ if you declare an
variable as:
const char i=90;
throught the program 'i' cant be modified
but #define macro can replace const.. and that macro can be
modified.... but its disadvantage is it will blindly
substitute the data which is #defined.......

Answer / vijay kumar (mainpuri)

Answer of #6 , output is wrong

correct out is given below :

Value is: 20
Value is: 20

(above output based on turbo c++ IDE)
Hence, const value can't be change.

Answer / sri harsha s

Its compiler dependant.
Using pointer you can modify the content of a 'const'
variable. It works in gcc.. you will get the following
warning when you compile that program.

warning: initialization discards qualifiers from pointer
target type.

I am not sure about turbo c++.

Answer / guest

use pointers and u can do anything...

Answer / mr.x

int main()
{
const volatile int no=10;
int *ptr;
ptr=(int *)&no;
*ptr=30;
printf("%d %d",no,*ptr);

return 0;
}

Answer / varunreddy

try this it will work:

#include<stdio.h>
int main()
{
const int i=10;

*(int *)&i=i++;
printf("%d\n",i);

return 0;
}

Answer / varunreddy

a small modification to answer #13

#include<stdio.h>
int main()
{
const int a=10;

*(int *)&a=25
printf("%d\n",a);
return 0;
}

Answer / raj

Answer 6 would be true if you change the variable to const
volatile.

The number of bytes of storage occupied by short, int and long are a) 2, 2 and 4 b) 2, 4 and 4 c) 4, 4 and 4 d) none

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