Returning address using function pointer

Asked Feb 01 '20 at 12:11

Active Feb 01 '20 at 12:11

Viewed 23 times

Hi All I have written a code as shown below

#include<iostream>
using std::cout;
using std::cin;
using std::endl;
int main()
{
    int *fun();
    int *p;
    p = fun();
    cout << "address contained in p " << p << endl;
    cout << "value contained  in p " << *p << endl;





    cin.get();

}
int *fun()
{
    int a = 20;
    return (&a);
}

In the output i get

address contained in p 00f3fc48

value contained in p 26238874

why is output different from what i expected (*p=20 as i am returning address of a from function pointer)??

I am writing this code in visual studio.

Does compiler matters in this case?

asked Feb 01 '20 at 12:11

rishi pandey

3

You are returning the address of a local variable declared only inside that function. That variable is destroyed once the function returns so this is not a good idea. – mathematician1975 Feb 01 '20 at 12:14
The *life-time* of all non-static local variables ends when the scope they were defined in end. For the variable `a` inside the function `fun`, its life-time ends when the function ends. – Some programmer dude Feb 01 '20 at 12:15
Thanks @Someprogrammerdude .Got the point. – rishi pandey Feb 01 '20 at 12:20

Returning address using function pointer

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