bool f(int &j) { return 0 = &j } Could the function f ever return true when called in a C++ program? If not, explain why. If so, give a code fragment which shows an example where f would return true. This was an interview question .. and this dumb me could not understand it

Well...are you sure it's supposed to say "return 0 = &j" and not "return 0 == &j" ?

if what you have written is correct, it shouldn't even compile...
you can't assign a value to 0.
However, if the other way is correct, then it would return true if the memory address of j was == to null. However, I don't know if you can have a variable w/ a memory address of 0x0, but I may be wrong.

If I am wrong on this, will someone please correct me?



/**************************************
* Hookt on Fonix relly werked fer mee!
* WarAmp
***************************************/

Edited by - Waramp on April 5, 2001 2:41:17 AM

It is theoretically possible, but it will never happen in practice.

I agree WarAmp. The code as is should produce an error that says something to the effect of 'can't assign &j to '0', lvalue is required'..

Testing in MSVC gives:

"cannot convert from 'class glExtManager *' to 'const int'. This conversion requires a reinterpret_cast, a C-style cast or function-style cast".

If I provide the C-style cast (like so: 'return 0 = (int)&j') then I get this:

"left operand must be l-value"

EDIT: If it is supposed to be 'return 0 == &j', then it should be impossible. Since the parameter definition for j is 'int& j' that means it is passed by reference. Checking the address of 'j' will _never_ be null because in C++ it is not possible to have a NULL reference.

--------------------------
I guess this is where most people put a famous quote...
"Everything is funnier with monkey''s" - Unknown

Edited by - Promiscuous Robot on April 5, 2001 2:59:17 AM

Yes. It can return true. What you do is you pass it a pointer. See my code, here.

main ()
{
int *q=new int;
q=NULL;

cout << f(*q) << endl;
}

bool f is already given.

Heh.. and I''m new to this. The reason this works is:

You are setting what q is pointing to to null. You pass the pointer with the *, you are talking about the item. The address of the item (which is non-existant) , is NULL. The comparison returns true, and life goes on.

Magnwa

Oh.. I was assuming that the very obvious lvalue thing was a mistype.

Magnwa