在下面的代码中,为什么s1.printVal会导致悬空指针错误？是不是s1对象,即它的指针,在它被销毁之前仍然可以访问？
class Sample
{
public:
int *ptr;
Sample(int i)
{
ptr = new int(i);
}
~Sample()
{
delete ptr;
}
void PrintVal()
{
cout << "The value is " << *ptr;
}
};
void SomeFunc(Sample x)
{
cout << "Say i am in someFunc " << endl;
}
int main()
{
Sample s1 = 10;
SomeFunc(s1);
s1.PrintVal(); // dangling pointer
}
解决方法
这里的问题是为SomeFunc()的参数完成的副本.该副本在销毁时取消分配指针.您还需要实现复制构造函数和复制赋值运算符.见
rule of three.
编辑：
这是“扩展”的伪代码,即编译器在main()函数中为您做的事情：
// main addr0 = grab_stack_space( sizeof( Sample )); // alloc stack space for s1 Sample::ctor( addr0,10 ); // call ctor of Sample addr1 = grab_stack_space( sizeof( Sample )); // alloc stack for argument Sample::ctor( addr1,addr0 ); // call COPY-ctor of Sample SomeFunc( addr1 ); // call SomeFunc Sample::dtor( addr1 ); // XXX: destruct the copy free_stack_space( addr1,sizeof( Sample )); // free stack taken by copy Sample::PrintVal( addr0 ); // call member func on s1 Sample::dtor( addr0 ); // destruct s1 free_stack_space( addr0,sizeof( Sample )); // YYY: free stack taken by s1
这不是确切的表示,而是概念性的解释.只需考虑编译器与代码有什么关系.
