Question:
How does one overload
operator->*()?
Answer:
Here is an example:
struct bar
{
void doSomething();
}
struct foo
{
bar *operator ->* () { return bar_; }
}
void Foo ()
{
void (bar::*barFn)() = &bar::doSomething;
foo f;
(f->*barFn)(); // calls bar::doSomething;
}
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