const vector implies const elements?

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难免孤独
难免孤独 2021-02-05 00:11
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  • 2021-02-05 00:27

    So a const object can only call const methods. That is:

    class V {
      public:
        void foo() { ... }        // Can't be called
        void bar() const  { ... } // Can be called
    };
    

    So let's look at a vector's operator[]:

    reference       operator[]( size_type pos );
    const_reference operator[]( size_type pos ) const;
    

    So when the vector object is const, it will return a const_reference.

    About: (*v[0]).set(1234);

    Let's break this down:

    A * const & ptr = v[0];
    A & val = *ptr;
    val.set(1234);
    

    Note that you have a constant pointer to variable data. So you can't change what is pointed at, but you can change the value that the pointer points at.

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  • 2021-02-05 00:35

    The first version

    v[0].set (1234); 
    

    does not compile because it tries to change the vector's first element returned to it by reference. The compiler thinks it's a change because set(int) is not marked const.

    The second version, on the other hand, only reads from the vector

    (*v[0]).set(1234);
    

    and calls set on the result of the dereference of a constant reference to a pointer that it gets back.

    When you call v[0] on a const vector, you get back a const reference to A. When element type is a pointer, calling set on it is OK. You could change the second example to

    v[0]->set(1234);
    

    and get the same result as before. This is because you get a reference to a pointer that is constant, but the item pointed to by that pointer is not constant.

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  • 2021-02-05 00:39

    Yes, a const vector provides access to its elements as if they were const, that is, it only gives you const references. In your second function, it's not the objects of type A that are const, but pointers to them. A pointer being const does not mean that the object the pointer is pointing to is const. To declare a pointer-to-const, use the type A const *.

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  • 2021-02-05 00:47

    Yes, because std::vector is a value-type rather than a reference type.

    To simplify things: An std::vector considers the values in its buffer as part of itself, so that changing them means changing the vector. This may be confusing if we only think of a vector as holding a pointer to an allocated buffer and the size: We don't change these two fields when we change elements in the buffer.

    It's the opposite than for pointers, which are reference-types; if you change the pointed-to value you haven't changed the pointer itself.

    The fact that std::vector is a value-type is a design choice - it's not something inherent in the C++ language. Thus, for example, the std::span class is also basically a pair of a pointer and a size, but an std::span can be const while you can still change the pointed-to elements. (There are other differences between spans and vectors.)

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