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cpointersconstants

Assigning address of a const variable to non const pointer

I have a question:

const int a = 10;
int *ptr;
ptr = (int *)&a;

What is the use of (int *) in the third line above?

Just like the above if we have:

char str[] = "abc";
char *pc;
pc = str;

Is the above assignment correct?

And if it is, then why is it not in the first case?

Solution 
  • const int a = 10;

    a is stored in read-only area (basically to tell compiler that this value cannot be changed).

    a is of type const int, and &a is of type const int *.

    So type casting is need to match it:

    int *ptr = (int *)&a;

    You can avoid it by declaring as below:

    int const *ptr = &a;

    p is of type const int, types are matching, no issue.

    updated for your query:

    Pointer to constant can be declared in the following two ways.

    1. const int *ptr;

    2. int const *ptr;

    Both are same. In this case we can change pointer to point to any other integer variable, but cannot change value of object (entity) pointed using pointer ptr.

    But this is not the same as 1st: int *const ptr; — this is a Constant pointer to a variable.
    In this case we can change value of object pointed by pointer, but cannot change the pointer to point another variable.