疑惑1:在类中,有参构造函数传入参数时,取地址符的效果
class Person { public: Person(int s, int a) { cout << "有参构造函数" << endl; sex = &s; age = &a; cout << *age <<" "<< *sex << endl; } ~Person() { cout << "析构函数" << endl; } int* sex; int* age; };
为什么输出时只能先输出p1.sex才能看到正确的值,且仅能看到p1.sex的值!