Local array scope

Hi,

When the method "foo" ends, shouldn't the arr[2] array be gone (due to
scope termination) and the pointer in class A be invalid? However,
when I run the program, I get "2". It seems to that I get this result
by luck; in further executions, I may get some other value. Am i
right?
class A
{
public:
A( int* ptr ){ arr = ptr; }
int* getArr(){ return arr; }
private:
int* arr;
};

A foo( )
{
int arr[2] = {2,1};
return A( arr );
}

int main()
{
A a = foo();
int* arr = a.getArr();

cout << arr[0] << endl; // This is not ambigous, since arr has
actually gone out of scope, am i right?
return 0;
}

Kai-Uwe Bux

Guest

n/a Posts

July 4th, 2008
12:45 PM
#2

Re: Local array scope

D. Susman wrote:

Quote:

Originally Posted by

Hi,
>
When the method "foo" ends, shouldn't the arr[2] array be gone (due to
scope termination) and the pointer in class A be invalid?

Yes.

Quote:

Originally Posted by

However, when I run the program, I get "2".

You dereference an invalid pointer. Anything can happen, including a value
of 2.

Quote:

Originally Posted by

It seems to that I get this result by luck; in further executions, I may
get some other value. Am i right?

Yes.

Quote:

Originally Posted by

class A
{
public:
A( int* ptr ){ arr = ptr; }
int* getArr(){ return arr; }
private:
int* arr;
};
>
A foo( )
{
int arr[2] = {2,1};
return A( arr );
}
>
int main()
{
A a = foo();
int* arr = a.getArr();
>
cout << arr[0] << endl; // This is not ambigous, since arr has
actually gone out of scope, am i right?

"Ambiguous" ????

It is undefined behavior. That's all.

Quote:

Originally Posted by

return 0;
}

Best

Kai-Uwe Bux