> in calling a Method in C++

Can you tell me the difference between a . and -> call to a method in C++.

This code works fine, using both calling methods.

#include <iostream>

using namespace std;

class myclass
{
    public:
    string doSomething();
};


string myclass::doSomething()
{
    return "done somethingn";
}

int main (int argc, const char * argv[])
{
    myclass c;
    std::cout << c.doSomething();

    myclass *c2;    
    std::cout << c2-&gt;doSomething();

    return 0;
}

I don't understand the different between the 2 calls? they both work?

---

The arrow operator is meant for calling a method from a pointer to an instance of an object.

The dot operator is meant for calling a method from a reference to an instance of an object, or on a locally defined object.

Your code would not compile if you reversed the operators on the two examples.

---

c2->doSomething();

is equivalent to:

(*c2).doSomething();

ie the pointer is being de-referenced before calling the method.

Check out Alf Steinbach's pointer tutorial for more help.

---

myclass *c2;
std::cout << c2->doSomething();

This is undefined behaviour. c2 is not initialized.

You needed to write

myclass *c2 = &c;
c2->doSomething();

c2->doSomething() is semantically equivalent to (*c2).doSomething() which in-turn is same as c.doSomething()

EDIT

Check out Alf Steinbach's pointer tutorial

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