violation of const variable using a pointer c++

Question

Why i am allowed to change a const value without errors?

void foo(long * a)
{
    *a = 50;
}

int _tmain(int argc, _TCHAR* argv[])
{
    const long a = 7; 
    const long * b = &a;

    foo((long *)b);
    // now the value of a is 50
}

Answer 1

You're not .

When you wrote that C-style cast, you disabled the checks. If you'd written this:

foo(static_cast<long*>(b));

or just

foo(b);

then your program would not have compiled.

In order to allow you to do expert magic, you can still write:

foo(const_cast<long*>(b));

but then it's still your responsibility to ensure that you do not use it for evil, such as the evil in your example.

Answer 2

Because you casted the constness away with the cast (long*) . Your program engenders undefined behavior as you attempt to modify a const object.

Especially when casting pointers you should rely on the C++-Casts ( static_cast , reinterpret_cast ), which do not implicitly cast away constness as the C-style cast does. That can only be done explicitly using const_cast , which is almost exclusively used for APIs with lacking const-correctness (and some hacks).
If you try to pass the argument without the cast (which is the way that arguments should usually be passed to functions):

foo(b);

You will get an appropriate error message .