C/C++ changing the value of a const

Question

I had an article, but I lost it. It showed and described a couple of C/C++ tricks that people should be careful. One of them interested me but now that I am trying to repli

Answer 1

In C++, Using Microsoft Visual Studio-2008

const int a = 3;    /* I promisse i won't change a */
int * ptr1  = const_cast<int*> (&a);
*ptr1 = 5;  /* I'm a liar, a is now 5 . It's not okay. */
cout << "a = " << a << "\n"; /* prints 3 */
int arr1[a]; /* arr1 is an array of 3 ints */

int temp = 2;
/* or, const volatile int temp = 2; */
const int b = temp + 1; /* I promisse i won't change b */
int * ptr2  = const_cast<int*> (&b);
*ptr2 = 5; /* I'm a liar, b is now 5 . It's okay. */
cout << "b = " << b << "\n"; /* prints 5 */
//int arr2[b]; /* Compilation error */

In C, a const variable can be modified through its pointer; however it is undefined behavior. A const variable can be never used as length in an array declaration.

In C++, if a const variable is initialized with a pure constant expression, then its value cannot be modified through its pointer even after try to modify, otherwise a const variable can be modified through its pointer.

A pure integral const variable can be used as length in an array declaration, if its value is greater than 0.

A pure constant expression consists of the following operands.

1. A numeric literal (constant ) e.g. 2, 10.53

2. A symbolic constant defined by #define directive

3. An Enumeration constant

4. A pure const variable i.e. a const variable which is itself initialized with a pure constant expression.

5. Non-const variables or volatile variables are not allowed.

Answer 2

this will create a runtime fault. Because the int is static. Unhandled exception. Access violation writing location 0x00035834.

void main(void)
{
    static const int x = 5;
    int *p = (int *)x;
    *p = 99;                //here it will trigger the fault at run time
}

Answer 3

we can change the const variable value by the following code :

const int x=5; 

printf("\nValue of x=%d",x);

*(int *)&x=7;

printf("\nNew value of x=%d",x);

Answer 4

#include<stdio.h>
#include<stdlib.h>

int main(void) {
    const int a = 1; //a is constant
    fprintf(stdout,"%d\n",a);//prints 1
    int* a_ptr = &a;
    *a_ptr = 4;//memory leak in c(value of a changed)
    fprintf(stdout,"%d",a);//prints 4
return 0;
}

Answer 5

You probably want to use const_cast:

int *ptr = const_cast<int*>(ptr_to_a);

I'm not 100% certain this will work though, I'm a bit rusty at C/C++ :-)

Some readup for const_cast: http://msdn.microsoft.com/en-us/library/bz6at95h(VS.80).aspx

Answer 6

The article you were looking at might have been talking about the difference between

const int *pciCantChangeTarget;
const int ci = 37;
pciCantChangeTarget = &ci; // works fine
*pciCantChangeTarget = 3; // compile error

and

int nFirst = 1;
int const *cpiCantChangePointerValue = &nFirst;
int nSecond = 968;

*pciCantChangePointerValue = 402; // works
cpiCantChangePointerValue = &ci; // compile error

Or so I recall-- I don't have anything but Java tools here, so can't test :)