Question

I want to transfer few variables using structure. Following is the sample program code. When I run this program, I get segmentation fault. I use gcc compiler.

Can anyone help me with this?

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

The problem is here:

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

You have not allocated memory for all. When you have an uninitialized pointer, it is pointing to random locations in the memory, which you probably won’t have access to. You have two options:

1. Allocate on the stack:

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

2. Allocate on the heap:

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

The first case is good when you know that all will be needed only in the current function (and those you call). Therefore, allocating it on the stack is sufficient.

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.