CODEWITHSUNDEEP
c
Pedro Javier Nicolas
Pedro Javier Nicolas

Reputation: 63

Memory in a linked list

I have been programming a simple game in C to learn programming, but I found an error in my code when I work with a linked list.

Here is my struct:

    typedef struct ListShoots * ListShoots;

    struct ListShoots{
        Shoot d;
        ListShoots next;
    };

And my functions about memory:

    ListShoots Shoots_Create(int x, int y){
        ListShoots ld = malloc(sizeof(struct ListShoots));
        ld->d.alto = 10;
        ld->d.x = x;
        ld->d.y = y;
        ld->d.v = 6;
        ld->next = NULL;
        return ld;
    }

    void Shoots_Shoot(ListShoots ld, int x, int y){
        ListShoots aux = ld;
        if(ld==NULL) printf("ld = NULL\n");
        while(ld!=NULL) ld = ld->next;
        ld = Shoots_Create(x,y);
        ld = aux;
        if(ld==NULL) printf("ld = NULL (again)\n");
    }

When I call the function Shoots_Shoot in the main (at first, ld = NULL), I get the output: ld = NULL and ld = NULL (again) too. Why? There should be just the first printf, I think, because at first the line while(ld!=NULL) ld = ld->next does nothing.

The weird thing is: If I remove the line ld = aux;, the output is okay, just: ld = NULL.

EDIT: The problem is: aux and ld are pointing at the same memory because of aux = ld. So if I reserve memory for ld, immediately aux changes and will be not pointing to NULL, right?

Upvotes: 1

Views: 83

Answers (3)

Shondeslitch
Shondeslitch

Reputation: 1079

- aux and ld are pointing to the same memory here:

 ListShoots aux = ld;

- But now if this loop is done at least once they not are pointing to the same memory:

while(ld!=NULL) ld = ld->next;

- Now ld points to another space memory due to malloc, while aux still points to the same place:

 ld = Shoots_Create(x,y);

- And you remember that you said that aux was pointing NULL at the beginning? Well, now ld are pointing to the same memory that aux again. It means NULL:

ld = aux;

Aux is always the same value (the value of ld when the method starts), so both conditionals test always print the same.

Upvotes: 0

mathematician1975
mathematician1975

Reputation: 21351

In regard to the specific problem you are describing look at your function

void Shoots_Shoot(ListShoots ld, int x, int y){
    ListShoots aux = ld;
    if(ld==NULL) printf("ld = NULL\n");
    while(ld!=NULL) ld = ld->next;
    ld = Shoots_Create(x,y);
    ld = aux;
    if(ld==NULL) printf("ld = NULL (again)\n");
}

The first line stores the passed in value of ld in aux variable. You then print if ld is null.

Later you change ld 

ld = Shoots_Create(x,y);

but immediately after (before the printf) you do so again back to its original value with the next line

ld = aux;

so essentially both conditionals are testing the same value (the value of ld passed to the function). This is why when you remove the line

ld = aux;

it works - because then you don't change the return value of Shoots_Create(x,y) function

Upvotes: 0

Beta Carotin
Beta Carotin

Reputation: 1679

... I get the output: "ld = NULL" and "ld = NULL(again)" too. ¿Why?

Because of

ListShoots aux = ld;
/* ... */
ld = aux;

You are restoring the original value of ld, which apparently is NULL, since you are getting

ld = NULL

from 

if(ld==NULL) printf("ld = NULL\n");

Upvotes: 2

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