Is using const_cast for read-only access to a const object allowed?

Question

In C++ I have a function that only requires read-only access to an array but is mistakenly declared as receiving a non-const pointer:

size_t countZeroes( int* array, size_t count )
{
    size_t result = 0;        
    for( size_t i = 0; i < count; i++ ) {
       if( array[i] == 0 ) {
           ++result;
       }
    }
    return result;
}

and I need to call it for a const array:

static const int Array[] = { 10, 20, 0, 2};

countZeroes( const_cast<int*>( Array ), sizeof( Array ) / sizeof( Array[0] ) );

will this be undefined behaviour? If so - when will the program run into UB - when doing the const_cast and calling the functon or when accessing the array?

Answer 1

Yes, it is allowed (if dangerous!). It's the actual write to a const object that incurs undefined behaviour, not the cast itself (7.1.5.1/4 [dcl.type.cv]).

As the standard notes in 5.2.11/7 [expr.const.cast], depending on the type of the object an attempt to write through a pointer that is the result of casting away const may produce undefined behaviour.

Answer 2

Since your code does not modify the array, and you told the compiler you know what you are doing by using the const_cast, you will actually be OK. However, I believe you are technically invoking undefined behaviour. Best to get the function declaration fixed, or write, declare and use the const-safe version of it.

Answer 3

Yes, you can do that. No, it is not undefined behavior as long as the function truely does not try to write to the array.

Answer 4

The problem of const_cast is always the same -- it allows you to "break the rules", just like casting to and from void* -- sure you can do that, but the question is why should you?

In this case it's of course ok, but you should ask yourself why didn't you declare size_t countZeroes( const int* array, size_t count ) in the first place?

And as a general rule about const_cast:

1. It may produce hard to find bugs
2. You're throwing away the const-agreement with the compiler
3. Basically you're turning the language into a lower-level one.

Answer 5

Using const_cast on an object which is initially defined as const is UB, therefore the undefined behaviour comes about immediately at the point you call const_cast.