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Does const vector<A> mean that its elements are constas well?
In the code below,
v[0].set (1234); in void g ( const vector<A> & v ) produces the compiler error
const.cpp:28:3: error: member function 'set' not viable: 'this' argument has type 'const value_type' (aka 'const A'), but function is not marked const
Why?
But (*v[0]).set(1234); in void h ( const vector<A *> & v ) is OK for the compiler.
What's the difference between the versions?
// ........................................................... class A { private: int a; public: A (int a_) : a (a_) { } int get () const { return a; } void set (int a_) { a = a_; } }; // ........................................................... void g ( const vector<A> & v ) { cout << v[0].get(); v[0].set (1234); } // () // ........................................................... void h ( const vector<A *> & v ) { cout << (*v[0]).get(); (*v[0]).set(1234); } // () 
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A const vector will return a const reference to its elements via the [] operator . In the first case, you cannot change the value of a const int&. In the second case, you cannot change the value of a reference to a constant pointer, but you can change the value the pointer is pointed to. – Julian.
A const iterator points to an element of constant type which means the element which is being pointed to by a const_iterator can't be modified.
If the vector itself is declared const (as in const std::vector<T*>), then you can't modify the vector, but you can modify the objects. If the pointers are declared const (as in std::vector<const T*>), then you can modify the vector, but not the objects.
You can't put items into a const vector, the vectors state is the items it holds, and adding items to the vector modifies that state. If you want to append to a vector you must take in a non const ref.
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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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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