C++中类的方法的调用 . 和->的区别

Question

#include <iostream>
using namespace std;
class Person{
    public:
    Person(){
    };
    void speak(){
    cout<<"Helloworld"<<endl;
};
};
int main()
{
    Person p;
    p->speak(); //error
    p.speak();
    return 0;
}

为什么这里面会出现错误？也就是说在什么情况下使用 -> ?

如果把里面的代码换成：

Person *p;
    p->speak();//right   这里面是不是默认实例化了一个类？
    p.speak();//error
    return 0;

求解惑，求指导。

Answer 1

Basic syntax knowledge, -> is used for class instances of pointer types. In addition, in the second case, it can only pass the compilation, but it will definitely cause a segment fault when running. The reason is that the constructor will not be called at this time. When you use a pointer, you must create a new object. I won’t say more. Yeah, it’s very tiring to grab a phone at the airport and reply haha

Answer 2

The pointer type has an overloaded -> operator, so it is no problem to use -> for pointers, and the . operator for objects.

Of course, if the object is overloaded -> can also be used -> It depends on whether you have implemented it yourself. If not, it will definitely not work.

Answer 3

a->someMethod() is generally equivalent to (*a).someMethod()
. is used for instances, -> is used for pointers. . .

But only in general.