Thread: Unable to stop at the end of linked list

Code:

#include<stdio.h>
#include<stdlib.h>

struct list1{
	char *name;
	struct list1 *next;
};

struct list1* freemem1(struct list1 **node)
{
	struct list1 *temp;
	temp=*node;
	if(temp==NULL)
	{ 
		temp=temp->next;
		free(*node);
	}
	return temp;
}

void addname(char name[],struct list1 **node)
{
	struct list1 *x,*ptr;
	char *s = malloc( sizeof( name ) );
	x = malloc(sizeof(struct list1));
	strcpy( s, name);
	x->name=s;

	x->next=NULL;
	ptr=*node;

	if(*node==NULL)
		*node=x;
	else{
		while(ptr->next!=NULL)
			ptr=ptr->next;
		ptr->next=x;
	}
}

int main(void)
{
	struct list1 *nList,*ptr;
	FILE *fileptr;
	char name[BUFSIZ],dump;
	nList= NULL;

	fileptr=fopen("namelist.txt","r");
	if(fileptr==NULL)
		perror("Unable to open file");
	else{
		while(fscanf(fileptr,"%[^\n]\n",name,&dump)!=EOF)
		{
			printf("%s\n",name);
			addname(name,&nList);
		}
		for(ptr=nList;ptr;ptr=freemem1(&ptr))
		{
			printf("%s\n", ptr->name);
		}
	}
	return 0;
}

>strcpy( s, name);
You need to include <string.h>.

>if(temp==NULL)
This should test for inequality:

Code:

if (temp != NULL)

That way you dereference temp and move foward, then release the memory if the pointer is not null rather than the other way around where you simply return a pointer to the node.

...erm.... got the idea... malloc(name[BUFSIZ]) doesn't allocate enough blocks of memory...
but i still donno how to solve the problem... can you give me some guide

Try strdup

and don't forget to free the memory...

>Try strdup
strdup isn't standard, so you can't be sure that the OP has it. Though it's trivial to write:

Code:

#include <stdlib.h>
#include <string.h>

char *strdup(const char *str)
{
  char *dup = malloc(strlen(str) + 1);

  if (!dup) {
    return NULL;
  }
  strcpy(dup, str);

  return str;
}

Of course, the name strdup is invading the implementation's namespace because any function name that begins with str and is followed by a lower case letter is reserved for future additions to <string.h>.

thx...this is a nice function ^_^.. save me a lot of time

>return dup;
Hmm, I seem to be making more mistakes than usual. Maybe I'm losing my touch.

oic... when an array passed into a function... it will be treated as a pointer in the function... and the sizeof array only return the size of the pointer itself not the allocated memory of the string.. isit?

I seem to be explaining this quite a bit these days, but except as the direct operand to sizeof, address-of (&) or a string literal initializer, array names are converted to a pointer to the first element. Those three situations are where the object is used rather than the value. So in the following code:

Code:

#include <stdio.h>

void f(char passed[])
{
  printf("%lu\n", (unsigned long)sizeof passed);
}

int main(void)
{
  char direct[10];

  printf("%lu\n", (unsigned long)sizeof direct);
  f(direct);

  return 0;
}

sizeof direct results in the value of 10 being printed and sizeof passed results in the size of a pointer being printed (4 on my system). That's because passing an array as the argument to a function doesn't fit into any of the situations where an array is used in object context., so a pointer to the first element is passed.

....erm.. the pointer is the only passed in element right?

>....erm.. the pointer is the only passed in element right?
The pointer is the only argument passed, yes.