Share pointer with const qualifier

首先定義一個 shared_ptr<int> p1
接著宣告另一個 shared_ptr<const int> p2 並指向 p1
分別把值印出來

#include <iostream>
#include <memory>

int main() {
    std::shared_ptr<int> p1 = std::make_shared<int>(3);
    std::shared_ptr<const int> p2 = p1;
    std::cout << "*p1 = " << *p1 << std::endl;
    std::cout << "*p2 = " << *p2 << std::endl;
    return 0;
}

output

p1 = 3
p2 = 3

表示 const shared_ptr 可以指向 shared_ptr

接著將 p1 的值改為 5 並印出來

#include <iostream>
#include <memory>

int main() {
    std::shared_ptr<int> p1 = std::make_shared<int>(3);
    std::shared_ptr<const int> p2 = p1;
    std::cout << "p1 = " << *p1 << std::endl;
    std::cout << "p2 = " << *p2 << std::endl;
+   *p1 = 5;
+   std::cout << "p1 = " << *p1 << std::endl;
+   std::cout << "p2 = " << *p2 << std::endl;
    return 0;
}

output

p1 = 3
p2 = 3
p1 = 5
p2 = 5

可以發現 p2 的值也被改變，那麼 const 代表的意義是什麼呢？
嘗試更改 p2 為 6

#include <iostream>
#include <memory>

int main() {
    std::shared_ptr<int> p1 = std::make_shared<int>(3);
    std::shared_ptr<const int> p2 = p1;
    std::cout << "p1 = " << *p1 << std::endl;
    std::cout << "p2 = " << *p2 << std::endl;
    *p1 = 5;
    std::cout << "p1 = " << *p1 << std::endl;
    std::cout << "p2 = " << *p2 << std::endl;
+   *p2 = 6;
    return 0;
}

output

ERROR!
/tmp/VDwjEvjmMm.cpp: In function 'int main()':
/tmp/VDwjEvjmMm.cpp:12:9: error: assignment of read-only location '((std::__shared_ptr_access<const int, __gnu_cxx::_S_atomic, false, false>*)(& p2))->std::__shared_ptr_access<const int, __gnu_cxx::_S_atomic, false, false>::operator*()'
   12 |     *p2 = 6;
      |     ~~~~^~~

compiling failed!

Conclusion

• It is valid to assign non const shared_ptr to const shared_ptr
• Modify const shared_ptr would fail