Why doesn't this object get set to NULL after the function call?

Bob John Published at Dev

Bob John

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

struct Person{
    char* first_name;
    char* last_name;
};

struct Person* create_person(const char* f_name, const char* l_name)
{
    struct Person* new_person = malloc(sizeof(struct Person));
    new_person->first_name = malloc((1+strlen(f_name))*sizeof(char));
    new_person->last_name = malloc((1+strlen(l_name))*sizeof(char));

    strcpy(new_person->first_name, f_name);
    strcpy(new_person->last_name, l_name);

    return new_person;
}

void delete_person(struct Person* person)
{
    free(person->first_name);
    free(person->last_name);
    free(person);

    person->first_name = 0;
    person->last_name = 0;
    person = 0;

    if(person){
        printf("Person not deleted\n");
    }
    else{
        printf("Person deleted\n");
    }
}

int main()
{
    struct Person* person1 = create_person("John", "Doe");
    delete_person(person1);

    if(person1){
        printf("Person not deleted\n");
    }
    else{
        printf("Person deleted\n");
    }
}

Program output is:

Person deleted

Person not deleted

My question is: why doesn't person1 point to NULL? It seems like the memory is freed (because if I try to use it I get an error), but why is the pointer still pointing to something (apparently) valid? Why doesn't person = 0; get carried across the function call?

Jack

To parameter's value change be see outside your function (and not only locally, like you're seeing) you need to take a struct Person ** person (note **) instead of.

EDIT: note that you're accessing memory which you don't own anymore in:

person->first_name = 0;
person->last_name = 0;

Because you've already freed person

Try this:

void delete_person(struct Person ** person)
{
    struct Person *p = *person;

    free(p->first_name);
    free(p->last_name);
    free(*person);

    *person = 0;
}

And call using: