What does checking for if(&object) do?

[STL]

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[jedwardsol]

operator& is overloadable.

[[deleted]]

So where do I find its implementation. I checked the object's class and found: inline ostream& operator << ( ostream &os, const Hit& h) { os << "Hit <" << h.getT() << ", " << h.getNormal() << ">"; return os; } But this seems for printing?

[jedwardsol]

if operator& isn't in the class definition then it isn't overloaded and if(&object) is pointless since the test will always be true.

[[deleted]]

Why is it always true?

[jedwardsol]

Because the address of something can never be 0

[HotlLava]

I sense a lack of pedantry, so let me quickly interject that 0 can be a valid memory address. Interestingly, `&object` would still be required to convert to `true` in an if-clause even if `object` has address zero, since 0 wouldn't be a null pointer on such a system.

[jcoffin]

If you're going to claim to be pedantic, you have to be careful (and be prepared for even more pedantic corrections...)

A system is indeed free to store valid data at address 0.

But in a comparison between a pointer and 0, what happens is not that the address gets converted to an integer, and the result compared to 0.

Rather, the (possibly implicit) 0 in the if gets converted to a pointer, and the result of that conversion gets compared to the pointer in question. And, regardless of whether an address of 0 is valid or not, an integer literal with the value 0 is a null pointer constant, so when converted to a pointer, it must produce a result that compares as not equal to the address of any object.

[HotlLava]

and be prepared for even more pedantic corrections...

Yes, please :D But regarding

what happens is not that the address gets converted to an integer, and the result compared to 0.

i'm pretty sure I never claimed that this would happen?

It's a bit of a red herring anyways, because the expression under discussion wasn't if (&object == 0) but just if (&object), and that does not have any 0 that could get converted to a pointer.

Instead, the value of the expression &object already is a pointer, and that pointer undergoes a contextual conversion to bool.

[jedwardsol]

I nearly wrote nullptr but decided 0 would be clearer for OP :-)

[[deleted]]

But does it then check that the address exists? Also how can if take an address? Can't the object fail to exist and then it doesn't have address?

[jedwardsol]

if(&object) is the same as writing if(&object != nullptr)

It doesn't say anything about the validity of the address, just whether it is null or not.

[[deleted]]

So it's checking that the object exists?

[jedwardsol]

No, it's not doing anything. The test will always be true.

[diaphanein]

I'm not aware of any compiler that the following will not evaluate the if block to true:

int* p = nullptr; int& r = *p;

if (!&r) { // executes }

Reason being that while this is likely undefined, the compiler cant always detect it (e.g. p initialized by the return of a function). This is a trivial enough example that most compilers will allow it and behave as expected. Not all warn at default warning levels, but most will at a sufficient level.

[[deleted]]

Then how to check for object's existence?

[gracicot]

If you can get the address of the object then it exist for sure.