what's wrong? dangling pointer?

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?

You're making a pointer point to a stack variable which is destroyed and then dereferencing the pointer, producing undefined behaviour — Seth Carnegie, Jan 03 '12 at 15:52
Undefined behavior which means you can get the expected output (http://ideone.com/VMYzG) or some other result (http://codepad.org/OiN8n1L0) — UncleBens, Jan 03 '12 at 16:20

score 8 · Accepted Answer · answered Jan 03 '12 at 15:53

8

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.

answered Jan 03 '12 at 15:53

unwind

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so #1 should output a random number? – bitstore Jan 03 '12 at 15:58
2

@tinybit: You need to realize that it's quite pointless to reason about what "should" happen. It's **undefined behavior**. – unwind Jan 03 '12 at 15:59
@tinybit: It shouldn't do anything specific. "Anything could happen". – Oliver Charlesworth Jan 03 '12 at 16:00
I get it. #1 out 10 confused me, what I thout was right. Thank you! – bitstore Jan 03 '12 at 16:03
@tinybit 10 is a random number :) – Seth Carnegie Jan 03 '12 at 16:10

Mike Seymour · Answer 2 · 2012-01-03T15:59:08.290

This is indeed a dangling pointer.

You are assigning p to point to an automatic (local) object. Once fun has returned, the object no longer exists, and attempting to access it through p gives undefined behaviour.

If you're interested in why you observe that particular behaviour: on most platforms, the stack frame of fun will still exist until another function is called. So reading p for the first call to << is quite likely to find the old value of i. After calling <<, the old stack frame has most likely been overwritten, so reading p will find an arbitrary value. But none of this is behaviour you can rely on; accessing the dead object could cause a crash, or any other behaviour.

+1 for explaining possible causes for output. While UB can't be relied on, one needs to be able to recognize the effects. — UncleBens, Jan 03 '12 at 16:24

score 1 · Answer 3 · answered Jan 03 '12 at 15:53

1

Yes, p becomes a dangling pointer the moment fun() returns.

answered Jan 03 '12 at 15:53

jrok

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score 1 · Answer 4 · answered Jan 03 '12 at 15:54

1

You are saving a pointer to a variable that is out of scope. Thus, the behavior is undefined. It can print anything, or even crash your application. Or even make your computer explode.

answered Jan 03 '12 at 15:54

score 1 · Answer 5 · answered Jan 03 '12 at 15:54

1

Your function is returning a pointer to something that gets over-written:

int i = 10;
p = &i; // This line

Because i is a local variable.

answered Jan 03 '12 at 15:54

Peter K.

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what's wrong? dangling pointer?

5 Answers5