Swap arrays in linked list

Question

So, I have to swap data of two nodes. Here is the function i created:

void swap(struct st* a, struct st* b)
{
    struct st* temp;

    temp->lname = a->lname; //lname is an array of chars
    a->lname = b->lname;
    b->lname = temp->lname;

    temp->marks = a->marks; //marks is an array of doubles
    a->marks = b->marks;
    b->marks = temp->marks;
}

So i do not understand why an error occures. The error is following: E0137 expression must be a modifiable lvalue

Here is the struct st

#define N 20
#define M 5
struct st
{
    char lname[N];
    char fname[N];
    int birth[3];
    double marks[M];
    double avarage_mark;

    struct st* next;
};

Answer 1

You are trying to change where an array is when you do a->lname = b->lname;. That's not allowed. You would need to strcpy the string from one struct st to another. Also, struct st* temp; is an uninitialized pointer. It doesn't point at any allocated memory so the program has undefined behavior trying to use it.

Another option is to copy the whole struct st at once:

void swap(struct st* a, struct st* b)
{
    struct st temp = *a;  // initialize temp with *a
    *a = *b;
    *b = temp;

    // swap next pointers back
    b->next = a->next;
    a->next = temp->next;
}

Answer 2

This declaration

struct st* temp;

declares an uninitialized pointer with an indeterminate value. So dereferencing the pointer as

temp->lname = a->lname;

invokes undefined behavior.

Secondly arrays do not have the assignment operator. Arrays are non-modifiable lvalues. And this error message

E0137 expression must be a modifiable lvalue

means that you are trying to assign one array to another.

If you need to swap data members of the structure st for two nodes then the function can look the following way

void swap(struct st* a, struct st* b)
{
    struct st temp = *a;
    *a = *b;
    *b = temp;

    struct st *p = a->next;
    a->next = b->next;
    b->next = p;
}