Question

Possible Duplicate:
Returning the address of local or temporary variable
Can a local variable's memory be accessed outside its scope?

#include<iostream>
using namespace std;

int *p = NULL;

void 
fun(void){
    int i = 10;
    p = &i;
}

int 
main(void){
    fun();
    cout<<*p<<endl; //#1
    cout<<*p<<endl; //#2
    return 0;
}

I think #1 and #2 will output the same, but why #1 output 10 and #2 output a random number?

Answer 1

This is just undefined behavior. You’re working with a pointer to a local variable, after that variable has gone out of scope. Anything could happen.