C Programming - Const - Discussion

Here Const int x = 5 means the x variable is const we cannot change ok.

Also, const int *p means the pointer pointing to a variable is also constant. (so in line number 3, p = &x,

So, now it means a pointer pointing to a constant value which is also defined as constant )

const int*p ---->>> const *p (means it is pointing to a constant value, so we cannot change it.)

If you remove const int x = 5 ------>>>>int x =5, then also it gives an error bcoz this statement const int *p means it is pointing to a constant value simple.

So you cannot change it.

1)const x is declared as integer and assign the value 5.

2)const ptrx value declared as integer point.

3)ptrx=&x mean ptrx contain address of x which mean *ptrx contain x value 5.

4)then *ptrx=10 mean indirectly change the value of 5 know.

So its wrong.

If any wrong in above statement means please any one clear me.

const int *ptrx -> Pointer to the constant integer.So we can't modify the value present at ptrx.

However if we want to modify then type casting will help. That is *(int *)ptrx = 10; or *(int *)&x = 10.

The variable "x" of integer data type is declared as constant with assigning value 5.

ptrx is a pointer to the integer data type. Means, it stores integer values only.

You may correct or suggest me.

int main()
{
int x=5;
int *ptrx;
ptrx = &x;
*ptrx = 10;
printf("%d\n", x);
return 0;
}
To change the value of const variable x we have to use *(int *)&x = 10;

Hi Shiva,

I think,

If const variable will be located in Read Only section of the memory. Trying to modify it will result in runtime error. Signal will be sent to the process in Linux.

How we can change const variable by using *(int *)&x = 10; (mentioned in explanation). However it is not allowing to change from 5 to 10. But no error also raising.

Using pointers we can change any kind it of data.

Even if it is constant variable also. That is why there is no security in C.

A small correction in the Explanation Step 2 is needed: ptrx is not a constant, but a pointer to an integer constant.

This is an undefined behavior by casting away the const definition. However, it will not work in g++ 4.1+.