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    <p>As AndreyT pointed out, under these circumstances assignment (mostly) doesn't make a lot of sense. The problem is that <code>vector</code> (for one example) is kind of an exception to that rule.</p> <p>Logically, you copy an object into the <code>vector</code>, and sometime later you get back another copy of the original object. From a purely logical viewpoint, there's no assignment involved. The problem is that <code>vector</code> requires that the object be assignable anyway (actually, all C++ containers do). It's basically making an implementation detail (that somewhere in its code, it might assign the objects instead of copying them) part of the interface.</p> <p>There is no simple cure for this. Even defining your own assignment operator and using <code>const_cast</code> doesn't really fix the problem. It's perfectly safe to use <code>const_cast</code> when you get a <code>const</code> pointer or reference to an object that you know isn't actually defined to be <code>const</code>. In this case, however, the variable itself <em>is</em> defined to be <code>const</code> -- attempting to cast away the <code>const</code>ness and assign to it gives undefined behavior. In reality, it'll almost always work anyway (as long as it's not <code>static const</code> with an initializer that's known at compile time), but there's no guarantee of it.</p> <p>C++ 11 and newer add a few new twists to this situation. In particular, objects no longer <em>need</em> to be assignable to be stored in a vector (or other collections). It's sufficient that they be movable. That doesn't help in this particular case (it's no easier to move a <code>const</code> object than it is to assign it) but does make life substantially easier in some other cases (i.e., there are certainly types that are movable but not assignable/copyable).</p> <p>In this case, you <em>could</em> use a move rather than a copy by adding a level of indirection. If your create an "outer" and an "inner" object, with the <code>const</code> member in the inner object, and the outer object just containing a pointer to the inner:</p> <pre><code>struct outer { struct inner { const double coeff; }; inner *i; }; </code></pre> <p>...then when we create an instance of <code>outer</code>, we define an <code>inner</code> object to hold the <code>const</code> data. When we need to do an assignment, we do a typical move assignment: copy the pointer from the old object to the new one, and (probably) set the pointer in the old object to a nullptr, so when it's destroyed, it won't try to destroy the inner object.</p> <p>If you wanted to badly enough, you could use (sort of) the same technique in older versions of C++. You'd still use the outer/inner classes, but each assignment would allocate a whole new inner object, or you'd use something like a shared_ptr to let the outer instances share access to a single inner object, and clean it up when the last outer object is destroyed.</p> <p>It doesn't make any real difference, but at least for the assignment used in managing a vector, you'd only have two references to an <code>inner</code> while the <code>vector</code> was resizing itself (resizing is why a vector requires assignable to start with).</p>
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