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Home/ Questions/Q 7651627
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T11:33:18+00:00 2026-05-31T11:33:18+00:00

for some reason calling my function ‘delAll’ more than once will cause a invalid

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for some reason calling my function ‘delAll’ more than once will cause a invalid free error from Valgrind. I don’t understand why if I call this function the second time would cause the program to go into the while loop again even though it just “delAll” of the node

//p is a linked list with call

struct node{
char *str, int data, struct node *next;
}

//here’s the function I am having trouble with:

void delAll()
{

struct node *temp,*temp2;
temp=p;
while(temp!=NULL)
{

    temp2=temp;
    temp= temp->next;

    free(temp2->str);
    free(temp2);

}

}
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  1. Editorial Team
    Editorial Team
    2026-05-31T11:33:19+00:00Added an answer on May 31, 2026 at 11:33 am

    p is the pointer to your list, and right now it will still after the delAll call point to the (free’d) start of the list. I’d just do;

    p=NULL;
    

    …right after your while loop to set p to null (ie have the list properly cleared). That will prevent your delAll from trying to free all elements again.

    Of course that would depend on p not just being a temporary variable, I’m assuming it’s the real “start of the list” pointer.

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