隐式虚拟性传播的原因是什么? [英] What is the reason of implicit virtual-ness propagation?
问题描述
我使用C ++的时间只有2到3个月，最近我发现虚函数后面的标识符 final .直到今天，我相信省略虚拟会阻止虚拟性的传播，但是我错了.它会隐式传播.
I've only been working with C++ for 2~3 months and recently I found out about the identifier, final, that comes after a virtual function. To this day, I believed that omission of virtual will stop the propagation of virtualness but I was wrong. It implicitly propagates.
我的问题是这个.为什么允许隐式传播?为什么没有 virtual 使功能虚拟化，而没有 virtual 使功能虚拟化呢?在某些情况下更好吗?还是回到第一次引入虚拟的那一天?
My question is this. Why allow implicit propagation? Why can't an existence of virtual make a function virtual and absense of virtual make a function not virtual? Is is better in some circumstance? or Was it, back in the day when virtual was first introduced?
根据 Clifford's答案，甚至有一个编译器会在缺少虚拟时生成警告.
According to Clifford's answer, there even is a compiler that generates warning upon absense of virtual.
我希望上面的链接可以回答我的问题，但是没有.
I expected above link to answer my question but it doesn't.
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有一些关于询问此功能有用性的评论. 我认为虚拟功能上的 final 关键字是使功能虚拟化的原因.该函数不能再被覆盖，因此派生类必须重新声明一个函数是否具有相同的名称. 如果最终与去虚拟化不同，请帮助我理解它. 如果最终没什么不同，那么从引入 final 的事实就可以看出去虚拟化的有用性. 我同意强迫显式虚拟会产生错误，但是我很好奇是否还有其他原因.
There are comments about asking the usefulness of this feature. I think final keyword on virtual function is what devirtualizes the function. The function can no longer by overridden, so a derived class must re-declare a function whether it has a same name or not. If final is different from devirtualization, Help me to understand it. If final is not different, then usefulness of devirtualization is self-evident from the fact final was introduced. I agree that forcing explicit virtual will produce bugs, but I'm curious if there are other reasons.
推荐答案
关于为什么存在(或不存在)特定功能的答案通常很困难，因为这成为一个问题.猜测和意见.但是，简单的答案可能是最小惊讶原则.提出一个合理且可靠且可预测的方案很困难.
The answer as to why a particular feature exists (or doesn't) is usually rather difficult, as it becomes a matter of guessing and opinions. However, the simple answer could be the principle of least astonishment. Coming up with a scheme that makes sense and works reliably and predictably would be difficult.
将功能虚拟化"甚至意味着什么?如果在运行时正在对象上调用去虚拟化"函数，它将使用指针的静态类型代替吗?如果静态类型具有虚函数，而运行时类型则没有，那么会发生什么?
What would "devirtualizing" a function even mean? If, at runtime, you're calling a "devirtualized" function on an object, would it use the static type of the pointer instead? If the static type has a virtual function but the runtime type doesn't, what happens?
#include <iostream>
struct A { virtual void f() const { std::cout << "A"; } };
struct B : A { void f() const { std::cout << "B"; } };
struct C : B { virtual void f() const { std::cout << "C"; } };
struct D : C { void f() const { std::cout << "D"; } };
void f(const A& o) { o.f(); }
int main()
{
// "devirtualized" real C++
f(A{}); // "A" "A"
f(B{}); // "A" or "B"? "B"
f(C{}); // "C"? "C"
f(D{}); // oh god "D"
}
还有一个事实，就是对于大多数设计，虚拟函数必须在整个层次结构中保持虚拟.要求全部使用virtual会引入各种难以诊断的错误. C ++通常试图避开那些需要纪律才能正确使用的功能.
There's also the fact that for the vast majority of designs, a virtual function has to stay virtual in the whole hierarchy. Requiring virtual on all of them would introduce all sorts of bugs that would be very hard to diagnose. C++ usually tries to stay away from features that require discipline to get right.
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