Thread: why is the linked list not being loaded ?

So I was trying my hands on a linked list in c, although I think I have done all well, my list is not being loaded when I try to load it...can somone point me to where I'm going wrong? now I post the snippet.. when I try to print out the items in the list nothing is printed out and when I try to count the items I get a count of zero...so I'm surely doing something wrong ...

Code:

typedef struct n
{
  char name [100];
  //double price;
  struct  n * next;
} node;
void loadItemsIntoList(node * testa, char nome [] );


int main()
{
node * list = 0;


//random list loading..
 loadItemsIntoList(list, "fff");
 loadItemsIntoList(list, "ssss");
 loadItemsIntoList(list, "sdfsd");
 loadItemsIntoList(list, "lknou");


 for (node * curr = list; curr != 0; curr = curr->next)
 {
     printf("%s", curr->name);
 }
 //testing for length of list
 int count = 0;
 for(node * curr = list; curr != 0; curr = curr->next)
 {
   count += 1;
 }
printf("\nlength of node is %d ,", count );
return 0;
}


void loadItemsIntoList(node * testa, char nome [] /*, double prezzo */)
{
     node * temp = malloc(sizeof(node));
     strcpy(temp->name, nome);
     //temp->price = prezzo;
     temp->next = 0;
  if(testa == 0)  {  testa = temp; }
  else{
        node * curr = testa;
        while(curr->next != 0){curr = curr->next;}
        curr->next = temp;
     }
}

Originally Posted by Salem

loadItemsIntoList modifies testa, but that change isn't returned back to main.

You need to do something like
list = loadItemsIntoList(list, "fff");

please what do you mean by "that change isn't returned to main? it doesn't say anything and with "you need to do something like this" well what prevents me from doing like how I did...loadItemsIntoList(...) returns a void ...
can you please be clearer?

Originally Posted by Xupicor

All this boils down to the following example:

Code:

#include <stdio.h>

void foo(int i) {
    i = 42;
}

int main() {
    int data = 0;
    foo(data);
    printf("%d", data);
}

What should be the printed value?

More specifically to your code - in C there's only pass-by-value when it comes to function parameters. The so called "pass by reference" (pass by pointer) in C is a convention of making the function take a pointer as a parameter, but the value of a pointer - the address - is still passed by value. It's copied.

Note the last sentence - it's copied. So if you work on a local copy - the change isn't reflected at the original.

a bit strange to me coming from c++, can you tweak my code a bit to make it work? I mean to make the list load the values? it will be a good and better example to the theory you just gave, I will see the difference..
infact in the short example you gave, I responded of course foo should update data to 42...but I just compiled and run it and no..it was still printing 0....I'm shocked

just did more test and the same thing is happening in c++, I got it that is boils down to locality, in the case of foo after foo exist it cleans up everything so data remains the same in main, I WAS THINKING data should be able to be updated by calling foo...but may be not because of function space and every function'ss activation record...
now I have to do some more test, but one should be able to modify a pointer having it address irrespective of where it is situated respective of a function..or I'm I wrong..?

The so called "pass by reference" (pass by pointer) in C is a convention of making the function take a pointer as a parameter, but the value of a pointer - the address - is still passed by value. It's copied.

if the address is copied..then I should be able to modify the value the address is pointing to right? because the value the address is pointing to is always one...
now lets see how this new realisation can be applied to the work at hand...

so in C can't I do something like
voidloadItemsIntoList(node * &testa, charnome [] ) ; and pass the pointer by reference?

ok, can you write a short snippet? taking into consideration the linked code that I posted up there?