这似乎是一个基本问题，但我没有看到它被问到：
假设以下简单情况：
没有虚拟成员。
虚拟继承用于允许多个路径指向同一个基。
就访问最派生类的成员所需的时间而言，虚拟继承的代价是多少？特别是，如果有一个非零价格，它是否只与通过多条路径继承的成员或其他成员有关？
【问题讨论】：
标签： c++ inheritance virtual-inheritance
通过额外的间接访问虚拟基类的数据成员。
【讨论】：
就访问最派生类的成员所需的时间而言，虚拟继承的价格是多少？
一个偏移查找和一个添加（2 条指令和一个内存提取）
特别是，如果存在非零价格，它是否仅与通过多个路径继承的成员有关，还是也与其他成员有关？
是的，即使这样也不总是。如果编译器有足够的信息证明不需要通过间接访问，则可以在编译时随意短路查找。
最好明确说明何时会出现这种情况。 ——尼可波拉斯
先生说得好。
这里有一个例子来证明这一点。使用 -O2 和 -S 选项进行编译以查看优化效果。
#include <memory>
#include <string>
enum class proof {
base,
derived
};
// volatile forces the compiler to actually perform reads and writes to _proof
// Without this, if the compiler can prove that there is no side-effect of not performing the write,
// it can eliminate whole chunks of our test program!
volatile proof _proof;
struct base
{
virtual void foo() const {
_proof = proof::base;
}
virtual ~base() = default;
};
struct derived : base
{
void foo() const override {
_proof = proof::derived;
}
};
// factory function
std::unique_ptr<base> make_base(const std::string&name)
{
static const std::string _derived = "derived";
// only create a derived if the specified string contains
// "derived" - on my compiler this is enough to defeat the
// optimiser
if (name == _derived) {
return std::make_unique<derived>();
}
else {
return {};
}
}
auto main() -> int
{
// here the compiler is fully aware that p is pointing at a derived
auto p = std::make_unique<derived>();
// therefore the call to foo() is made directly (in fact, clang even inlines it)
p->foo();
// even here, the compiler 'knows' that b is pointing at a 'derived'
// so the call to foo is made directly (and indeed on my compiler, completely
// inlined)
auto b = std::unique_ptr<base>(new derived);
b->foo();
// here we assign a derived to b via indirect construction through a string.
// Unless the compiler is going to track this string and follow the logic in make_base
// (and on my compiler it does not) this will prevent the virtual call to foo() from
// being turned into a direct call.
// Therefore, this call will be made via the virtual function table of *b
b = make_base("derived");
if (b) {
b->foo();
}
return 0;
}
【讨论】：