传递char未初始化的指针地址以及成员变量的分配和提取 [英] Passing char uninitialiezed pointer address and allocation and extraction of member variables

问题描述

我正在将 char * 指针从main传递到函数 abc ，该函数是未初始化的指针.我正在组成一个网络数据包，就像在功能中分配的那样，我希望它是数据包的表示形式.所以这是我在 main

I am passing char * pointer from main to a function abc which is uninitialized pointer. And I am composing a network packet so like what allocated in the function I want it to be the representation of a packet. So This is my passing of uninitialized pointer in function call inside main

            char *pay;
            abc(&pay);

在我具有的功能中，如下所示的功能

in the function I have as signature of a function like following

            void abc(char **c)

这是全部功能

        void abc(char **c)
        {
            *c=(char *) malloc(sizeof(struct ethhdr)+sizeof(struct iphdr)+sizeof(struct tcphdr)+1000/*1000=payload*/);
            struct ethhdr *eth=(struct ethhdr *)&*c;
            struct iphdr *ip=(struct iphdr *)(&*eth+sizeof(struct ethhdr));
            struct tcphdr *tcp=(struct tcphdr *)(&*ip+sizeof(struct iphdr));
            //populate eth
            //...
            //populate ip
            //...
            //populate tcp
            tcp->source=80;
            printf("%d\n",tcp->source);
            int *x=(int *)*c;
            //x=10;

        }

我想知道如何在 abc 函数中转换第二行我就是这样

I like to know how can I cast second line in abc function I am doing like this

        struct ethhdr *eth=(struct ethhdr *)&*c;

abc 函数内的第二行.

但在包含此指令 tcp-> source = 80; 的行中，它会导致segFault.

But at line containing this instruction tcp->source=80; its causing segFault.

所以我的问题是如何在函数中强制转换 c 双指针并将其分配给 struct ethhdr * ， struct iphd * 和 struct tcphdr * 指针

So my question is how can I cast the c double pointer in function and to assign it tostruct ethhdr *,struct iphd * and struct tcphdr * pointers

以及如何将值分配给 struct ethhdr ， struct iphdr 和 struct tcphdr

And how to assign the values to members of struct ethhdr , struct iphdr and struct tcphdr

我尝试过

tcp-> source = 80;

tcp->source=80;

但是会导致segFault

But causing segFault

推荐答案

您有两个问题.

第一个是在以下位置同时使用address-of和dereference运算符:

The first is the use of both the address-of and dereference operators in:

struct ethhdr *eth=(struct ethhdr *)&*c;

表达式& * c 与普通的 c 相同，这是不正确的.您需要在这里 * c :

The expression &*c is the same as plain c, which isn't correct. You need *c here:

struct ethhdr *eth = (struct ethhdr *) *c;

您对其他指针(例如& * eth )也有类似的问题，但是它不会因此引起任何问题.例如，仅使用普通的 eth .

You have similar problem with the other pointers (e.g. &*eth) but there it doesn't cause any problems because of this. Use just plain eth (for example).

第二个问题是双重的，是所有其他分配.首先，您使用的指针不是字节指针，这意味着强制转换为另一个结构指针将破坏严格别名.

The second problem is two-fold, and is all the other assignments. First of all the pointers you use are not byte pointers which means casting to another structure pointer will break strict aliasing.

第二，您必须记住，对于任何指针或数组 p 和索引/偏移量 i ，表达式 *(p + i)完全等于 p [i] .对于类似 eth + sizeof(struct ethhdr)的表达式，这意味着它与&)eth [sizeof(struct ethhdr)])相同.显然这是错误的( eth 可以看作是一个元素的数组，而不是至少包含 sizeof(struct ethaddr)+ 1 个元素的数组.

Secondly you have to remember that for any pointer or array p and index/offset i, the expression *(p + i) is exactly equal to p[i]. For an expression like eth+sizeof(struct ethhdr) that means it's the same as &)eth[sizeof(struct ethhdr)]). Which is clearly wrong (eth could be seen as an array of one element, not an array of at least sizeof(struct ethaddr) + 1 elements.

要解决这两个问题，包括严格的混叠和偏移量，您需要使用原始的字节指针 * c 并将偏移量(以字节为单位)添加到其中:

To fix both these problems, strict aliasing and the offsets, you need to use the original byte-pointer *c and add the offsets in bytes to that:

struct iphdr *ip = (struct iphdr *) (*c + sizeof(struct ethhdr));
struct tcphdr *tcp = (struct tcphdr *) (*c + sizeof(struct ethhdr) + sizeof(struct iphdr));

---

从上面开始，现在应该相对容易地找出如何获取有效负载数据的指针，因为它不是您的代码中使用的普通 * c .

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