Consider the following code:
class B
{
int x;
public:
B() : x( 10 ) {}
int get_x() const { return x; }
void set_x( int value ) { x = value; }
};
class A
{
boost::shared_ptr<B> b_;
public:
boost::shared_ptr<B> get_b() const { return b_; } // (1)
};
void f( const A& a)
{
boost::shared_ptr<B> b = a.get_b();
int x = b->get_x();
b->set_x( ++x ); // (2)
}
int main()
{
A a;
f( a );
return 0;
}
In this code (2) compiles without any errors or warnings independently the fact that get_b
is a const function and a
is a const object.
My question is how do you deal with this situation? The best I could use is to change (1) to the following:
boost::shared_ptr<开发者_开发问答const B> get_b() const { return b_; } // (1)
But I should always remember that I should add const
to the return type. That's not very convenient. Is there a better way?
This doesn't actually have anything to do with shared pointers per se. I mean if you had a plain pointer you'd have exactly the same problem and would solve it in exactly the same way, that is
const B* get_b() const {return b_; }
If you left it like
B* get_b() const {return b_; }
you'd have the same problem.
Well, you have found the solution yourself.
boost::shared_ptr<const B> get_b() const { return b_; } // (1)
It's the only const-correct way to do it.
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