the second printf, identical to my first printf, makes the program crash - c

Question

i'm obviously not blaming printf here, i probably messed up my mem allocations and access but i can't understand where i did wrong. the program crash on the second printf in main. It also crash on the third if i comment the second one. Actually it crash whenever i access p after the first printf !
Can someone explains me what i am doing wrong ?
Thanks a lot.
typedef struct
{
char * firstname;
char * lastname;
int age;
} person;
person * new_person(char * firstname, char * lastname, int age)
{
person p;
int lf = strlen(firstname);
int ll = strlen(lastname);
p.firstname = (char *)malloc(++lf * sizeof(char));
p.lastname = (char *)malloc(++ll * sizeof(char));
strcpy(p.firstname, firstname);
strcpy(p.lastname, lastname);
p.age = age;
return &p;
}
int main()
{
person * p = new_person("firstname", "last", 28);
printf("nom : %s ; prenom : %s ; age : %d\n", p->lastname, p->firstname, p->age);
printf("nom : %s ; prenom : %s ; age : %d\n", p->lastname, p->firstname, p->age);
printf("nom : %s ; prenom : %s ; age : %d\n", (*p).lastname, (*p).firstname,(*p).age);
return 0;
}

Answer 1

You're returning the address of a local variable.
You can either modify your new_person to take an argument (a pointer to a person) or you can malloc one inside the function and manipulate that.
The person you declared in your function goes out of scope when the function returns. Everything that happens to it after that is undefined. It might coincidentally keep its value for a while, but you should not depend on this. When you make your call to printf, the stack grows and overwrites the old location of your person with new stuff.

Answer 2

I think the issue is in this line:
return &p;
Notice that you are returning a pointer to a local variable. This results in undefined behavior, since as soon as the function returns, the local variable p no longer exists. As a result, reading or writing that pointer will read or write garbage data.
The fact that this doesn't immediately crash is an artifact of how the compiler generates code. Chances are, the first time you call printf, it reuses space that was previously used for p in a way that, by sheer coincidence, works out fine. However, after the function returns, its stack frame has clobbered the old memory for p. As a result, the second call to printf is reading garbage data left behind from the call to printf, hence the crash.
(Specifically: when you pass the parameters, it copies the pointers to the strings onto the stack, so when printf runs, it's probably trashing the original pointers, but using the copies. The second call then loads garbage pointers from the expired printf stack frame, since it lives at the same address that p used to live in.)
To fix this, consider changing p to a pointer to a person and then using malloc to allocate it. That way, the memory persists beyond the function call, so this crash should go away.
Hope this helps!

Answer 3

person p;
// stuff
return &p
This is wrong. After the function returns, the local variable is leaving the scope - its address will be invalid. You have to allocate the structure on the heap:
person *new_person(char *firstname, char *lastname, int age)
{
person *p = malloc(sizeof(*p));
p->firstname = strdup(firstname);
p->lastname = strdup(lastname);
p->age = age;
return p;
}

Answer 4

The problem is in the function new_person. You create person p on the stack and return its address. You need to allocate person* p = new person(.....