没有显式指针的C ++多态调用 [英] C++ polymorphic call without explicit pointer
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问题描述
如果我有基础课:
struct Base
{
void foo()
{
bar();
}
virtual void bar()
{
}
};
和派生类:
struct Derived : public Base
{
void bar()
{
cerr << "Derived here\n";
}
};
有可能在编写此代码时:
It is happen that when write this code:
Derived d;
d.foo();
由于调用了Derived::bar
,因此我将看到打印派生于此处".但是我不是通过指向base的指针来调用的,而是在这里工作的多态性.为什么?是因为Base::foo
中对bar
的调用实际上是对this->bar()
的隐式调用,而bar
是在类的vtable中找到的?
I will see printing "Derived here" - since Derived::bar
was called. But I did not call via pointer to base, but polymorphism working here. Why? Is it because the call to bar
in Base::foo
is implicitly actually to this->bar()
and bar
is finding in vtable of class?
推荐答案
您的猜测是正确的(尽管请记住,C ++标准对vtable没有任何说明).
Your guess is precisely correct (although bear in mind that the C++ standard says nothing about vtables).
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