c - 使用结构传递变量

Question

我想使用结构传输一些变量。以下是示例程序代码。当我运行这个程序时，我得到了分段错误。我使用 gcc 编译器。

谁能帮我这个？

struct data{
    const char * ip;
    const char * address;
    const char * name;
};

int fun(struct data *arg) {
    //struct data *all;
    const char *name = arg->name;
    printf("\n My name is:%s",name);
    return 1; 
}


int main(int argc, char * const argv[]) {
    struct data * all;   
    int k=0;
    //data.name = argv[1];

    all->name=argv[1];
    k = fun(all);

    printf("\n k is:%d ",k);

    return 0;
}

Answer 1

问题在这里：

struct data * all;
all->name=argv[1];

您尚未为all. 当您有一个未初始化的指针时，它指向内存中的随机位置，您可能无法访问这些位置。你有两个选择：

1. 在栈上分配：

   struct data all;
   all.name=argv[1];
   k = fun(&all);
   

2. 在堆上分配：

   struct data *all = malloc(sizeof(*all));
   if (all != NULL)
   {
       all->name=argv[1];
       k = fun(all);
   }
   free(all);
   

当您知道all仅在当前函数（以及您调用的函数）中需要时，第一种情况很好。因此，将其分配到堆栈上就足够了。

The second case is good for when you need all outside the function creating it, for example when you are returning it. Imagine a function initializing all and return it for others to use. In such a case, you can't allocate it on the stack, since it will get destroyed after the function returns.

You can read more about it in this question.

Answer 2

You need to allocate memory and assign it to the pointer all:

struct data * all;   
int k=0;
all = malloc(struct data);

all->name=argv[1];
k = fun(all);
//...
free(all);

or use a local struct and pass its pointer to the function:

struct data all;   
int k=0;

all.name=argv[1];
k = fun(&all);

Answer 3

all is a pointer to type struct data. You never assigned it, so it doesn't point to anything.

You need to do one of the following:

• allocate all on the stack:

  struct data all;

• allocate all on the heap:

    struct data* all = malloc(sizeof (struct data));   
    // don't forget to check if the allocation succeeded, 
    //and don't forget to free it when you're done