sockaddr 和 sockaddr_in 之间的转换

问题描述

我遇到了一个引用它的套接字编程教程

I came across a socket programming tutorial in which it is quoted

指向 struct sockaddr_in 的指针可以转换为指向 struct sockaddr 的指针反之亦然"

我不明白如何将 sockaddr_in 转换为 sockaddr.将 Big 类型的指针转​​换为 Small 类型应该会产生 UD 行为.

I dont understand how can sockaddr_in be cast to sockaddr. Casting a pointer of Big type to Small type should give UD behavior.

struct sockaddr {
unsigned short sa_family; // address family, AF_xxx
char sa_data[14]; // 14 bytes of protocol address
};


struct sockaddr_in {
short int sin_family; // Address family, AF_INET
unsigned short int sin_port; // Port number
struct in_addr sin_addr; // Internet address
unsigned char sin_zero[8]; // Same size as struct sockaddr
};

演员阵容如何不被定义?将这些相互投射会不会不安全?

How can the cast not be undefined? Isn't it unsafe to cast these to each other?

如果我有一个只有两个整数的 A 类和有 4 个整数的 B 类.如果我有一个 B 类型的指针并且我将它转换为 A 类型，那么确定我可以获取前两个元素.但是，如果 A 类首先声明了 2 个字符，然后声明了 2 个整数，那么指针将无法正确获取值，因为在这种情况下对象布局会有所不同.

If i have a class A having only two ints and class B having 4 ints. And if i have a pointer of type B and i cast it to type A then sure i can fetch the first two elements. But if class A has 2 chars declared first and 2 ints declared later then the pointers would not right fetch the values since the object layout in this case would be different.

编辑 1:

class Anu
{
public:
    char a;
    int b;
    Anu()
    {
        a='a';
    }
};
class Anurag
{
public:
    Anurag() { a=4;}
    int a;
    int b;
    int c;
    int d;
};
int main()
{
        Anu objanu;
        Anurag objanurag;
        Anurag *ptrAnurag= &objanurag;
        ptrAnurag= (Anurag*)&objanu;
        cout<<ptrAnurag->a; //Some weird value here
        return 0;
}

假设我通过调整变量类型更改示例，使两个类具有相同的大小...即使大小保持不变，对象布局仍可能不同.

Assuming i change the example so that both classes have same size by adjusting the variables types...still the object layout might be different even though the size remains the same.

推荐答案

我会在@gsamaras 的回答中补充说，未定义的行为并不总是意味着坏事即将发生.未定义的行为实际上是说我们*没有提供任何关于如果发生 XYZ 时接下来会发生什么的规范".

I'll add to @gsamaras answer by saying that Undefined Behaviour doesn't always means that bad things are about to happen. Undefined Behaviour actually says "we* don't provide any specifications on what should happen next if XYZ occurs".

(*C++ 标准).

这是操作系统发生并说它由我们定义"的地方.

this is the place where the OS takes place and say "it is defined by us".

尽管强制转换不相关的结构(sockaddr_in,sockaddr)可能是标准未定义的行为，但 OS API 指定它对其 API 有效.

although casting unrelated structs (sockaddr_in,sockaddr) may be undefined behaviour by the standard, the OS API specify that it is valid with their API.

这篇关于sockaddr 和 sockaddr_in 之间的转换的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！