Why does std::vector transfer its constness to the contained objects?

Question

A const int * and an int *const are very different. Similarly with const std::auto_ptr<int> vs. std::auto_ptr<const int>. However, there appears to be no such distinction with const std::vector<int> vs. std::vector<const int> (actually I'm not sure the second is even allowed). Why is this?

Sometimes I have a function which I want to pass a reference to a vector. The function shouldn't modify the vector itself (eg. no push_back()), but it wants to modify each of the contained values (say, increment them). Similarly, I might want a function to only change the vector structure but not modify any of its existing contents (though this would be odd). This kind of thing is possible with std::auto_ptr (for example), but because std::vector::front() (for example) is defined as

const T &front() const;
T &front();

rather than just

T &front() const;

There's no way to express this.

Examples of what I want to do:

//create a (non-modifiable) auto_ptr containing a (modifiable) int
const std::auto_ptr<int> a(new int(3));

//this works and makes sense - changing the value pointed to, not the pointer itself
*a = 4;
//this is an error, as it should be
a.reset();


//create a (non-modifiable) vector containing a (modifiable) int
const std::vector<int> v(1, 3);

//this makes sense to me but doesn't work - trying to change the value in the vector, not the vector itself
v.front() = 4;
//this is an error, as it should be
v.clear();

Answer 1

It's a design decision.

If you have a const container, it usually stands to reason that you don't want anybody to modify the elements that it contains, which are an intrinsic part of it. That the container completely "owns" these elements "solidifies the bond", if you will.

This is in contrast to the historic, more lower-level "container" implementations (i.e. raw arrays) which are more hands-off. As you quite rightly say, there is a big difference between int const* and int * const. But standard containers simply choose to pass the constness on.

Answer 2

The difference is that pointers to int do not own the ints that they point to, whereas a vector<int> does own the contained ints. A vector<int> can be conceptualised as a struct with int members, where the number of members just happens to be variable.

If you want to create a function that can modify the values contained in the vector but not the vector itself then you should design the function to accept iterator arguments.

Example:

void setAllToOne(std::vector<int>::iterator begin, std::vector<int>::iterator end)
{
    std::for_each(begin, end, [](int& elem) { elem = 1; });
}

If you can afford to put the desired functionality in a header, then it can be made generic as:

template<typename OutputIterator>
void setAllToOne(OutputIterator begin, OutputIterator end)
{
    typedef typename iterator_traits<OutputIterator>::reference ref;
    std::for_each(begin, end, [](ref elem) { elem = 1; });
}