scanf with const int* const works, but shouldn't

Question

I have code equivalent to the following:

const int* const n = new int;
printf("input: ");
scanf("%d", n);
delete n;

Now, since n is a pointer to a CONSTANT integer, this shouldn't work (I'm expecting a compiler error). However, this seems to work properly and even stores the value of the input into *n.

I want to know, why doesn't this give me an error; why does it work? Shouldn't scanf be unable to alter the value of *n?

Answer 1

The prototype for scanf is:

int scanf ( const char * format, ... );

The ellipsis means it takes variable arguments. C (and C++) have no way to check the validity of these parameters. That's why you won't get an error if you pass the address of a double here or even the double variable itself. It's up to the programmer to verify the correct parameter is passed.

Answer 2

scanf has next to no type safety, it merrily does what you tell it to. This is because of the way variable-argument lists are implemented in C. They expect the types to be of the kind which you tell it.

So if you give scanf a conversion specifier which doesn't match, you will invoke undefined behavior, which occurs in run-time. Similarly, there is probably no way for the compiler to tell that the pointer passed is of type const type* const. A good compiler may give a diagnostic if it spots something fishy, but is by no means required to do so.

Since most cases of undefined behavior occur in run-time, it is generally the programmer's responsibility to know about the various forms of undefined behavior and avoid them.