从 directive_pointer_cast
获得返回值是否正常? 还是会引起问题?
struct A
{
};
struct B : public A
{
};
int main()
{
shared_ptr<A> b = make_shared<B>();
auto &a = dynamic_pointer_cast<A>(b);
//auto a = dynamic_pointer_cast<A>(b);
return 0;
}
从 directive_pointer_cast
获得返回值是否正常? 还是会引起问题?
struct A
{
};
struct B : public A
{
};
int main()
{
shared_ptr<A> b = make_shared<B>();
auto &a = dynamic_pointer_cast<A>(b);
//auto a = dynamic_pointer_cast<A>(b);
return 0;
}
即使它 are possible, 它也不会做你想做的事:
[目的是]参考柜台没有增加
由于 liversal_pointer_cast
返回一个 new shared_ptr
副本, 仍然会增加 。
您的代码不起作用, 因为此引用将会被 dyanamic_pointer_cast
所返回的临时物体约束, 而此选项是被禁止的 。
使用 const
引用将有效( 因为 const&
允许与临时约束), 但结果还是相同的: 投影将创建 < em> new em > 实例 < code > 共享_ ptr , 并增加引用数 。
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