This is my code
I am expecting this line should print 2 but I don't get any valueCode:#include<stdio.h> #include<stdlib.h> struct node { int data; struct node * next; }; void ADD ( struct node *Head, int value) { struct node * new = malloc( sizeof(*new)); if ( new != NULL ) { new->data = value; new->next = NULL; } if ( Head == NULL ) { Head = new; } } int main () { struct node * head = NULL; ADD(head,2); printf("%d", head-> data ); return 0; }
Code:printf("%d", head-> data );
Because you don't return the pointer from the function.
Do you understand why this doesn't print 42?
Code:void change(int foo) { foo = 42; } int main ( ) { int x = 0; change(x); printf("Result=%d\n", x); }
You need to start paying attention to your previous threads.
There are about a dozen threads from you on just lists.
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Yes, because we are only passing copy of variableOriginally Posted by Salem
I am having hard time to understand linked list. I am reading my old threads many time and trying to understand them
Can it be written in another way that doesn't require returing pointer from functionCode:#include<stdio.h> #include<stdlib.h> struct node { int data; struct node * next; }; struct node * ADD ( struct node *Head, int value) { struct node * new = malloc( sizeof(*new)); if ( new != NULL ) { new->data = value; new->next = NULL; } if ( Head == NULL ) { Head = new; } return Head; } int main () { struct node * head = NULL; head = ADD(head,2); printf("%d", head-> data ); return 0; }
The key is to understand why do we return a pointer from the ADD function in the first place. The answer to that is that we want to update the head pointer from the caller, and we cannot do that within the function as we only have a copy of that head pointer. Hence, we return the updated pointer so that the caller can assign it to the head pointer from the caller, thereby updating it.
So, what other ways can we use to update the head pointer from the caller? One common method is to pass a pointer to the head pointer, e.g.,
Note that you didn't implement the code to handle what happens when the head pointer is not a null pointer. I would also keep things simple by handling malloc returning a null pointer by returning early, rather than having to think about whether it is safe for new to be a null pointer further down in the function. For example:Code:void ADD(struct node **Head, int value) { struct node *new = malloc(sizeof(*new)); if (new != NULL) { new->data = value; new->next = NULL; } if (*Head == NULL) { *Head = new; } }
This also opens the possibility of returning an error code, whereas if you designed this function to return the head pointer, then you might need another pointer parameter to record an error code that the caller can examine.Code:void ADD(struct node **Head, int value) { struct node *new = malloc(sizeof(*new)); if (!new) { return; } // from here on you know that new is not a null pointer... new->data = value; new->next = NULL; if (*Head == NULL) { *Head = new; } }
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