Difference between `const shared_ptr` and `shared_ptr`?

问题

I'm writing an accessor method for a shared pointer in C++ that goes something like this:

class Foo {
public:
    return_type getBar() const {
        return m_bar;
    }

private:
    boost::shared_ptr<Bar> m_bar;
}

So to support the const-ness of getBar() the return type should be a boost::shared_ptr that prevents modification of the Bar it points to. My guess is that shared_ptr<const Bar> is the type I want to return to do that, whereas const shared_ptr<Bar> would prevent reassignment of the pointer itself to point to a different Bar but allow modification of the Bar that it points to... However, I'm not sure. I'd appreciate it if someone who knows for sure could either confirm this, or correct me if I got it wrong. Thanks!

回答1:

You are right. shared_ptr<const T> p; is similar to const T * p; (or, equivalently, T const * p;), that is, the pointed object is const whereas const shared_ptr<T> p; is similar to T* const p; which means that p is const. In summary:

shared_ptr<T> p;             ---> T * p;                                    : nothing is const
const shared_ptr<T> p;       ---> T * const p;                              : p is const
shared_ptr<const T> p;       ---> const T * p;       <=> T const * p;       : *p is const
const shared_ptr<const T> p; ---> const T * const p; <=> T const * const p; : p and *p are const.

The same holds for weak_ptr and unique_ptr.

回答2:

boost::shared_ptr<Bar const> prevents modification of the Bar object through the shared pointer. As a return value, the const in boost::shared_ptr<Bar> const means that you cannot call a non-const function on the returned temporary; if it were for a real pointer (e.g. Bar* const), it would be completely ignored.

In general, even here, the usual rules apply: const modifies what precedes it: in boost::shared_ptr<Bar const>, the Bar; in boost::shared_ptr<Bar> const, it's the instantiation (the expression boost::shared_ptr<Bar> which is const.

回答3:

#Check this simple code to understand... copy-paste the below code to check on any c++11 compiler

#include <memory>
using namespace std;

class A {
    public:
        int a = 5;
};

shared_ptr<A> f1() {
    const shared_ptr<A> sA(new A);
    shared_ptr<A> sA2(new A);
    sA = sA2; // compile-error
    return sA;
}

shared_ptr<A> f2() {
    shared_ptr<const A> sA(new A);
    sA->a = 4; // compile-error
    return sA;
}

int main(int argc, char** argv) {
    f1();
    f2();
    return 0;
}

回答4:

I would like to a simple demostration based on @Cassio Neri's answer:

#include <memory>

int main(){
    std::shared_ptr<int> i = std::make_shared<int>(1);
    std::shared_ptr<int const> ci;

    // i = ci; // compile error
    ci = i;
    std::cout << *i << "\t" << *ci << std::endl; // both will be 1

    *i = 2;
    std::cout << *i << "\t" << *ci << std::endl; // both will be 2

    i = std::make_shared<int>(3);
    std::cout << *i << "\t" << *ci << std::endl; // only *i has changed

    // *ci = 20; // compile error
    ci = std::make_shared<int>(5);
    std::cout << *i << "\t" << *ci << std::endl; // only *ci has changed

}

来源：https://stackoverflow.com/questions/17793333/difference-between-const-shared-ptrt-and-shared-ptrconst-t

标签

c++