small linked list programme in c, random printf output - c

Question

Im writing a little school programme, I have to use 'void const *content' as a parameter.
I'm having trouble printing the content of the new node. without 'const' the code works and displays everything correct. could someone point out what I'm doing wrong?
terminal output:
�
6
#include <stdio.h>
#include <stdlib.h>
typedef struct s_list
{
void *content;
size_t content_size;
struct s_list *next;
} t_list;
t_list *lstnew(void const *content, size_t content_size)
{
struct s_list *new = (struct s_list*)malloc(sizeof(struct s_list*));
if(new == NULL){
printf("No allocation!");
exit(1);
}
new->content = &content;
new->content_size = content_size;
new->next = NULL;
return(new);
}
int main(void)
{
printf("%s\n", lstnew("Hello", 6)->content);
printf("%zu\n", lstnew("Hello", 6)->content_size);
return(0);
}

Answer 1

You're taking the address of a local variable here:
new->content = &content;
Instead, just take the value:
new->content = content;
Also, you don't allocate enough memory here; you're only allocating enough for the pointer instead of the size of the structure:
struct s_list *new = (struct s_list*)malloc(sizeof(struct s_list*));
The cast on malloc is also unnecessary. I would write it like this:
struct s_list *new = malloc(sizeof(*new));
Instead of using a typedef and t_list, you should just use struct s_list everywhere, because the structure is not intended to be opaque.

Answer 2

In this declaration
struct s_list *new = (struct s_list*)malloc(sizeof(struct s_list*));
instead of allocation memory for an object of the type struct s_list there is being allocated memory for pointer of the type struct s_list *.
You have to write either
struct s_list *new = malloc( sizeof( struct s_list ) );
or
t_list *new = malloc( sizeof( t_list ) );
In this statement
new->content = &content;
the left-side operand has the type void * while the right-hand operand has the type const void **. Moreover you are using a pointer to the local variable content (function parameters are its local variables) that will not be alive after exiting the function.
What you need is to allocate memory and copy the content of the variable content in the allocated memory.
Here is a demonstrative program that shows how the function can be defined.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef struct s_list
{
void *content;
size_t content_size;
struct s_list *next;
} t_list;
t_list * lstnew( const void *content, size_t content_size )
{
t_list *new = malloc( sizeof( t_list ) );
if ( new != NULL )
{
new->content = malloc( content_size );
if ( new->content == NULL )
{
free( new );
new = NULL;
}
else
{
memcpy( new->content, content, content_size );
new->content_size = content_size;
new->next = NULL;
}
}
return new;
}
int main(void)
{
t_list *head = lstnew( "Hello", 6 );
head->next = lstnew("World!", 7 );
printf( "%s %s\n", ( char * )head->content, ( char * )head->next->content );
return 0;
}
The program output is
Hello World!