libpcap 捕获的包长度可能大于 MTU 吗？
其实是有可能的，关键在于网卡的 Offload，简单来说操作系统为了更好的 IO 性能，可能几个包一起交付给网卡然后硬件分片/分段，然而 libpcap 工作在操作系统和网卡之间所以虽然最后发出去的包小于 MTU 但是 libpcap 看到的是大于 MTU 的啦。
其实是有可能的，关键在于网卡的 Offload，简单来说操作系统为了更好的 IO 性能，可能几个包一起交付给网卡然后硬件分片/分段，然而 libpcap 工作在操作系统和网卡之间所以虽然最后发出去的包小于 MTU 但是 libpcap 看到的是大于 MTU 的啦。
struct base1
{
virtual ~base1() { }
int value1;
};
struct base2
{
virtual ~base2() { }
int value2;
};
struct derived : public base1, public base2
{
};
int main()
{
derived obj;
base1* p1 = &obj;
base2* p2 = &obj;
assert(static_cast<void*>(p1) == static_cast<void*>(p2));
}
C++ is amazing!好吧，其实就是多继承问题。
How2Code
struct base1 { virtual ~base1() { } int value1; }; struct base2 { virtual ~base2() { } int value2; }; struct derived : public base1, public base2 { }; int main() { derived obj; base1* p1 = &obj; base2* p2 = &obj; assert(static_cast<void*>(p1)…
struct base1More amazing.
{
virtual ~base1() { }
int value1;
};
struct base2
{
virtual ~base2() { }
int value2;
};
struct derived : public base1, public base2
{
};
int main()
{
derived obj;
base1* p1 = &obj;
base2* p2 = &obj;
assert(dynamic_cast<void*>(p1) == dynamic_cast<void*>(p2));
}
原因是 dynamic_cast 返回的地址是最派生类的起始地址，也就是 &obj.
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