Linked List Delete Node , Simple linked list - c

Question

I'm trying to implement a function that deletes nodes from a linked list. So far, I can delete just the first node of the list(3).
I tried to go to the for loop from delete, I thought that the memory is not well allocated, I have been struggling for a few days and I don't understand, please help me a little, it's the topic I received from college.
#include <stdio.h>
#include <stdlib.h>
typedef struct nod
{
int key;
struct nod *urm;
} NOD;
NOD *first=0,*last=0;
void add(int x)
{
NOD *p=(NOD*)malloc(sizeof(NOD));
p->key=x;
p->urm=0;
if(0==first)
{
first=p;
last=p;
}
else{
last->urm=p;
last=p;
}
}
void delete(int x)
{
NOD *q,*p;
if(first->key==x)
{
p=first;
first=first->urm;
free(p);
}
else{
for(p=q=first;p=0;q=p,p=p->urm)
{
if(p->key==x)
{
q->urm=p->urm;
if(p==last)
{
last=q;
}
free(p);
}
}
}
}
void show()
{
for(NOD *p=first;p!=0;p=p->urm)
{
printf("%d ",p->key);
}
printf("\n");
}
int main()
{
add(3);
add(1);
add(2);
add(5);
show();
delete(2);
show();
return 0;
}

Answer 1

For starters the code you showed is not a C++ code. It is a C code.
It is a bad idea to define global variables like first and last and when functions depend on global variables. In this case you can not create more than one list in a program.
As for the function delete then in general it has undefined behavior. It can be called for an empty list.
Moreover in this ;loop
for(p=q=first;p=0;q=p,p=p->urm)
there is a typo in the condition expression. You are using the assignment operator instead of the comparison operator.
And you function ignore the case when the list contains only one node because in this case it does not update the last node.
Nevertheless using your approach the function delete can look the following way.
void delete(int x)
{
if ( first )
{
if ( first->key == x )
{
NOD *tmp = first;
first = first->urm;
free( tmp );
if ( first == NULL ) last = NULL;
}
else
{
NOD *p = first;
while ( p->urm != NULL && p->urm->key != x )
{
p = p->urm;
}
if ( p->urm != NULL )
{
NOD *tmp = p->urm;
p->urm = p->urm->urm;
free( tmp );
if ( p->urm == NULL ) last = p;
}
}
}
}
Here is a demonstrative program.
#include <stdio.h>
#include <stdlib.h>
typedef struct nod
{
int key;
struct nod *urm;
} NOD;
NOD *first=0,*last=0;
void add(int x)
{
NOD *p=(NOD*)malloc(sizeof(NOD));
p->key=x;
p->urm=0;
if(0==first)
{
first=p;
last=p;
}
else{
last->urm=p;
last=p;
}
}
void delete(int x)
{
if ( first )
{
if ( first->key == x )
{
NOD *tmp = first;
first = first->urm;
free( tmp );
if ( first == NULL ) last = NULL;
}
else
{
NOD *p = first;
while ( p->urm != NULL && p->urm->key != x )
{
p = p->urm;
}
if ( p->urm != NULL )
{
NOD *tmp = p->urm;
p->urm = p->urm->urm;
free( tmp );
if ( p->urm == NULL ) last = p;
}
}
}
}
void show()
{
for(NOD *p=first;p!=0;p=p->urm)
{
printf("%d ",p->key);
}
printf("\n");
}
int main()
{
add(10);
add(20);
add(30);
add(40);
show();
delete(30);
show();
add( 50 );
add( 60 );
add( 70 );
add( 80 );
show();
delete(80);
show();
return 0;
}
Its output is
10 20 30 40
10 20 40
10 20 40 50 60 70 80
10 20 40 50 60 70

Answer 2

You can greatly simplify deleting a node from your list by using both a pointer-to-pointer to NOD and a pointer-to NOD to iterate over the list. This allows you to set the node at the current address to the node->urm eliminating the need to keep track of the previous node in the list, e.g.
/** delete node with key v from list (for loop) */
void delete (int v)
{
NOD **ppn = &first; /* pointer to pointer to first*/
NOD *pn = first; /* pointer to first */
for (; pn; ppn = &pn->urm, pn = pn->urm) {
if (pn->key == v) {
*ppn = pn->urm; /* set node at address to next node */
free (pn); /* free deleted node */
break;
}
}
}
See Linus on Understanding Pointers for further discussion on the advantages of using the address of a node in addition to a pointer to node.

Answer 3

I think your for loop condition is incorrect inside delete functions:
for(q=first, p=first->urm; p!=0; q=p, p=p->urm)
just change the condition, then it should work.