Is it Undefined Behaviour to cast away the constness of a function parameter?

Question

Imagine I have this C function (and the corresponding prototype in a header file)

void clearstring(const char *data) {
    char *dst = (char *)data;
    *dst = 0;
}

Is there Undefined Behaviour in the above code, casting the const away, or is it just a terribly bad programming practice?

Suppose there are no const-qualified objects used

char name[] = "pmg";
clearstring(name);

Answer 1

The attempt to write to *dst is UB if the caller passes you a pointer to a const object, or a pointer to a string literal.

But if the caller passes you a pointer to data that in fact is mutable, then behavior is defined. Creating a const char* that points to a modifiable char doesn't make that char immutable.

So:

char c;
clearstring(&c);    // OK, sets c to 0
char *p = malloc(100);
if (p) {
    clearstring(p); // OK, p now points to an empty string
    free(p);
}
const char d = 0;
clearstring(&d);    // UB
clearstring("foo"); // UB

That is, your function is extremely ill-advised, because it is so easy for a caller to cause UB. But it is in fact possible to use it with defined behavior.