syntax error of linked lists in dynamic memory in C language

Question

void push(struct node* head,int data)
{
    struct node* newnode=(struct node*)malloc(sizeof(struct node)); 
    newnode->data=data;
    newnode->next=head;
    head=newnode; //it doesnt work
}

This code is not working; the error that occured was not a syntax error.

So is it a syntax error or not a syntax error? You need to declare the parameter as `struct node** head` and change the last statement to `*head=newnode`. — Alex Shesterov, Mar 29 '16 at 21:38
the change to `head` won't be visible outside the `push` function, and I presume that's exactly what you'd like. — isedev, Mar 29 '16 at 21:38
code is not working means a lot, please point out your error. — Dilip Kumar, Mar 29 '16 at 21:38
the code is taken from a book..where the writer assumes that this code doesnt work — 77981, Mar 29 '16 at 21:40
@77981 All function parameters are passed by value in C. So the `head` variable within the `push` function is not the same `head` variable that the caller passed in (it's just the same value). So changing one does not affect the other. The variable needs to be passed in as `struct node **head;` and set with `*head=newnode;`. — kaylum, Mar 29 '16 at 21:43
A syntax error means it doesn't compile, and as a result you can't even attempt to run it. It sounds instead like this is a functional error. The last statement, where you assign to `head`, is dead code since it has no effect after the function returns. `head` is just a function argument, which is basically a type of local variable that is bound upon entry to the function. You probably intended to pass a *pointer* to head so that you can change it in the caller. — Tom Karzes, Mar 29 '16 at 21:43
see [C: change pointer passed by value](http://stackoverflow.com/questions/5531765/c-change-pointer-passed-by-value) — isedev, Mar 29 '16 at 21:44
if the function is not void,and it returns the variable head,does it work ? — 77981, Mar 29 '16 at 21:48

score 0 · Answer 1 · edited May 23 '17 at 12:15

Parameters in C are passed by value, rather than passed by reference, that is, formal parameters are just copies of actual parameters. Maybe you can understand why the following function doesn't work:

void negate(int num)
{
    num = -num; // Doesn't work
}

The reason why the posted code doesn't work is just similar. head = newnode; only changes the value of formal parameter head. After push() returns, the value of actual parameters from the caller remain unmodified. Thus, for the caller, nothing happens(expect that a block of memory is malloc()ed but not free()ed).

To change the value of variables, you must pass their address to functions:

void push(struct node **head,int data)
{
    struct node* newnode = malloc(sizeof(struct node)); 
    newnode->data = data;
    newnode->next = head;
    *head = newnode;
}

See also: How do I modify a pointer that has been passed into a function in C?

syntax error of linked lists in dynamic memory in C language

1 Answers1