struct and free() in C - c

Question

My data gets overwritten whenever I insert a new value ,If I omit my free() in the main program my program works fine.Why ?How to recify this issue.Is memory allocation of structure is correct?
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
struct List
{
char val[20] ;
};
struct Hashtable
{
struct List *start;
};
struct Hashtable ht[26];
void init();
void insert(struct List*);
void init()
{
register int j;
for (j=0;j<26;j++)
{
ht[j].start=NULL;
}
}
int main(void)
{
init();
int i=0;
for (int i=0;i<5;i++)
{
struct List *newnode=(struct List*)malloc(sizeof(struct List));
scanf("%s",newnode->val);
insert(newnode);
free(newnode);
newnode=NULL;
}
return 0;
}
void insert(struct List *node)
{
if ( ht[node->val[0]-97].start==NULL)
{
ht[node->val[0]-97].start=node;
return;
}
else
{
printf("The value is %s\n", ht[node->val[0]-97].start->val);
}
}
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Answer 1

When you assign pointers, you only copy the pointers themselves, not the memory they might point to.
When you call free using a pointer, all copies of that pointer become invalid and can no longer be used.
You either need to make a completely new copy in the insert function (including allocating a new List structure), or you should not call free.
My personal recommendation is that you don't allocate at all in the main function, and that the insert function takes the string to "insert" as an argument (instead of the List pointer it currently takes).
Perhaps something like this:
void insert(char *val)
{
// Get the hash-table index (note that it only works with ASCII encoding)
char hash = tolower(val[0]) - 'a';
// First check if it exists
if (ht[hash].start == NULL)
{
// No, then add it
// First allocate memory for the node
struct List *node = malloc(sizeof *node);
// Then copy the string
strcpy(node->val, val);
// And finally add it
ht[hash].start = node;
}
}
Then your loop in the main function could be like
for (unsigned i = 0; i < 5; ++i)
{
char val[20];
scanf("%19s", val);
insert(val);
}

Answer 2

Imagine memory is a row of lockers, each containig an 8 bit value, and numbered as 1, 2, 3... (Well, higher values usually, but you get it). A pointer is just the number of that locker, nothing more. What malloc does, is:
Find a spot with free memory
Mark that memory as used
Return the number of the first locker
When you free, you mark again that memory as free. There are many algorithms to do this, but it would be wise to search for a small free memory space whith its number as low as possible, since leaving empty spaces is a waste of memory.
When you assign a pointer, you do not assign the values it points to, but the locker number.
When you free, you give the algorithm the possibility to use that locker, and when you use malloc again, it probably finds the same locker as before, and then you modify its contents.
Remember that your hashtable still uses that locker number, and when it opens the locker, it will find out its contents are modified.
Solution below:
TL;DR:
Your hash table uses still the same memory, so don't free it, since it might be modified when found by malloc.
Free it from the hashtable once you are finished using that pointer, to avoid wasting memory
Note: If you just read the solution, understanding heap (above), will help you with a lot of headaches, plus it is nice to know what the computer is doing with your code.