这一定是一个错误。构造的本地对象的析构函数foo()
在接收对象的移动构造函数之前调用。尤其,似乎临时已分配但未(移动)构造按值返回时。下面的程序展示了这一点:
#include <iostream>
using namespace std;
struct C
{
C(int z) { id = z; cout << "C():" << id << endl; }
~C() { cout << "~C():" << id << endl; }
C(const C& c) { id = c.id + 1; cout << "C(const C&):" << id << endl; }
C& operator=(const C&) { cout << "operator=(const C&)\n"; return *this; }
C(C&& c) { id = c.id + 1; cout << "C(C&&):" << id << endl;}
C& operator=(C&&) { cout << "operator=(C&&)\n"; return *this; }
int id;
};
C foo() { C c(10); return c; }
int main()
{
const C c = foo();
return 0;
}
Output:
C():10
// THE TEMPORARY OBJECT IS PROBABLY ALLOCATED BUT *NOT CONSTRUCTED* HERE...
~C():10 // DESTRUCTOR CALLED BEFORE ANY OTHER OBJECT IS CONSTRUCTED!
C(C&&):4198993
~C():4198992
~C():4198993
创造two里面的物体foo()
似乎对这个问题有了更多的了解:
C foo() { C c(10); C d(14); return c; }
Output:
C():10
C():14
~C():14
// HERE, THE CONSTRUCTOR OF THE TEMPORARY SHOULD BE INVOKED!
~C():10
C(C&&):1 // THE OBJECT IN main() IS CONSTRUCTED FROM A NON-CONSTRUCTED TEMPORARY
~C():0 // THE NON-CONSTRUCTED TEMPORARY IS BEING DESTROYED HERE
~C():1
有趣的是,这似乎取决于对象的构造方式foo()
. If foo()
是这样写的:
C foo() { C c(10); return c; }
然后出现错误。如果这样写,则不会:
C foo() { return C(10); }
最后定义的输出foo()
:
C():10 // CONSTRUCTION OF LOCAL OBJECT
C(C&&):11 // CONSTRUCTION OF TEMPORARY
~C():10 // DESTRUCTION OF LOCAL OBJECT
C(C&&):12 // CONSTRUCTION OF RECEIVING OBJECT
~C():11 // DESTRUCTION OF TEMPORARY
~C():12 // DESTRUCTION OF RECEIVING OBJECT