Re: What three steps in overloading the arrow operator?
On Sun, 27 Jan 2013 13:25:54 -0800, fl wrote:
<snip>
In other words, we want to call the result of evaluating point->action.
The compiler evaluates this code as follows:
1. If point is a pointer to a class object that has a member named
action, then the compiler writes code to call the action member of that
object.
2. Otherwise, if point is an object of a class that defines operator->,
then point->action is the same as point.operator->()->action, That is,
we execute operator->() on point and then repeat these three steps,
using the result of executing operator-> on point.
3. Otherwise, the code is in error."
I do not know the last part of item 2. I do not find any snippet code
using ->display. I do not find any part mentioning 3 steps. Thus, I
cannot solve this myself. Could you help me out? Thanks,
The three steps mentioned in item 2 are the three numbered items I quoted
above.
The operator-> is special, because it's application can be repeated
without this being directly observable in the code.
For example:
#include <iostream>
struct Foo {
void print() { std::cout << "Hello from Foo" << std::endl; }
};
struct Bar {
Foo mFoo;
Foo* operator->() { return &mFoo; }
};
struct Baz {
Bar mBar;
Bar& operator->() { return mBar; }
};
int main()
{
Foo aFoo, *pFoo = &aFoo;
Bar aBar;
Baz aBaz;
pFoo->print(); /* Item 1: print is a member of Foo */
aBar->print(); /* Item 2: use operator-> */
aBaz->print(); /* Item 2 repeated: use operator-> twice */
}
This code compiles and prints "Hello from Foo" three times.
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