[Solved] *head’ is a pointer; did you mean to use ‘->’? Why am I getting this

Question

On this line:

*head->prev=temp;

The -> operator has higher precedence than the * operator, so it parses as:

*(head->prev)=temp;

This is invalid because head is a pointer-to-pointer-to-struct, not a pointer-to-struct. You need to add parenthesis to force the * operator to apply directly to head:

(*head)->prev=temp;

Also, don’t cast the return value of malloc as it can mask other errors in your code.

Accepted Answer

On this line:

*head->prev=temp;

The -> operator has higher precedence than the * operator, so it parses as:

*(head->prev)=temp;

This is invalid because head is a pointer-to-pointer-to-struct, not a pointer-to-struct. You need to add parenthesis to force the * operator to apply directly to head:

(*head)->prev=temp;

Also, don’t cast the return value of malloc as it can mask other errors in your code.