Linked list doubt previous year question

Question

The following C function takes a single-linked list of integers as a parameter and rearranges the elements of the list. The function is called with the list containing the integers 1, 2, 3, 4, 5, 6, 7 in the given order. What will be the contents of the list after function completes execution?

struct node {
    int value;
    struct node *next;
};

void rearrange(struct node *list) {
    struct node *p, *q;
    int temp;
   1: if (!list || !list -> next) return;
    p = list; q = list -> next;
    while(q) {
        temp = p -> value; p->value =  q -> value;
        q->value = temp; p = q ->next;
        q = p? p ->next : 0;
    }   
}

What is meant by line 1?

Comments

Either list pointer is null, or list has only one element so that "next" of initial pointer will be null.

Answer 1

@hem chandraJoshi It means that if  the list is empty or the list has only 1 element then return.

!list is another way of saying list==NULL, both will return 1 if list is NULL. (negation of NULL is non-NULL, which in C is described as some non-zero value, and in most compilers is taken as 1, so !NULL is 1.)

!list->next is another way of saying !(list->next), as -> has higher priority. This condition checks that the next pointer of first node is NULL.

Comments

There will be two conditions

1. List is empty

2. List has item : 1,2,3,4

As now we reach line 1

Case 1. List is empty  ! list = ! (null) = if(true) return ;

Case 2: List is not empty ! (list ) = ! (not null ) = if (null ) return ;=>if(false) return ;

Is it right interpretation ? If not plz make it right .

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Yes, it's right. When condition evaluates to null if code will not be executed.

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answer will be 2143657 ternary operator doesnt fail but q becomes NULL at the end