Question

Hey I am getting this error:

error: conversion to non-scalar type requested

Here are my structs:

typedef struct value_t value;

struct value{
   void* x;
   int y;
   value* next;
   value* prev;
};

typedef struct key_t key;

struct key{
   int x;
   value * values;
   key* next;
   key* prev;
};

Here is the code that is giving me problems:

struct key new_node = (struct key) calloc(1, sizeof(struct key));
struct key* curr_node = head;

new_node.key = new_key;

struct value head_value = (struct value) calloc(1, sizeof(struct value))

Am I not suppose to use calloc on structs? Also, I have a struct that I have created and then I want to set that to a pointer of that same struct type but getting an error. This is an example of what I am doing:

struct value x;
struct value* y = *x;

this gives me this error

error: invalid type argument of ‘unary *’

When I do y = x, I get this warning:

warning: assignment from incompatible pointer type

Answer 1

You are trying to assign a pointer expression (the return type of malloc() and friends is void*) to a struct type (struct new_node). That is nonsense. Also: the cast is not needed (and possibly dangerous, since it can hide errors)

struct key *new_node = calloc(1, sizeof *new_node);

the same problem with the other malloc() line:

struct value *head_value = calloc(1, sizeof *head_value);

More errors: You are omitting the ‘struct’ keyword (which is allowed in C++, but nonsense in C):

struct key{
   int x;
   struct value *values;
   struct key *next;
   struct key *prev;
};

UPDATE: using structs and pointers to struct.

struct key the_struct;
struct key other_struct;
struct key *the_pointer;

the_pointer = &other_struct; // a pointer should point to something

the_struct.x = 42;
the_pointer->x = the_struct.x; 

 /* a->b can be seen as shorthand for (*a).b :: */
(*thepointer).x = the_struct.x;

/* and for the pointer members :: */

the_struct.next = the_pointer;
the_pointer->next = malloc (sizeof *the_pointer->next);