c – 不允许带=符号的shared_ptr

template< class Y >
explicit shared_ptr( Y* ptr );

这样,您就无法使用复制初始化(shared_ptr< parent> p2 = new parent;).

我能想到为什么std :: shared_ptr的构造函数是明确的最好的原因是你不太可能犯错误将std :: shared_ptr对象的原始非托管传递给接收std :: shared_ptr的函数.

#include <memory>


void func(std::shared_ptr<int> ptr) { /* ... */}

int main() {
    auto iptr = std::make_shared<int>();
    func(iptr);  // OK

    /*
    int* iptr = new int;
    func(iptr);  // Unclear who would eventually manage (delete) the pointer
    // ...
    delete iptr; // Undefined behavior!
    */
}

如果没有明确构造函数,那么您可以执行注释掉的代码.

另外,就像@R一样. Martinho Fernandes有suggested,如果传递非新手指针,你可能会遇到问题.

int myint;
int* iptr = &myint;
func(iptr);  // iptr doesn't point to a newed (dynamically-allocated) object
             //   std::shared_ptr,by default,uses delete on its managed pointer
             //   if there are no more references to that pointer

如果你在使用两个单独的std :: shared_ptrs管理原始指针时不小心,你仍然可以做错.

int* riptr = new int;
std::shared_ptr<int> iptr(riptr);
std::shared_ptr<int> iptr2(riptr); // WRONG!

关于原始指针的问题,两者基本相同,但不一定适用于std :: shared_ptr.