int 0 cast to a pointer =? null pointer?

[MusCihOz]

Is a run-time integral value of 0, cast to a pointer, guaranteed to be a null pointer?

[Athar]

No, you only get that guarantee for a constant expression of value 0.

[webJose]

Really? I would have imagined it was. If not, how can this work?

char *ptr = new (std::nothrow) char[100];
if (!ptr) { cout << "Out of memory"; }

After all, I am comparing the pointer there to a value. It is the equivalent of if (ptr == 0). I can also write a function that returns a pointer, and I can return 0 to signify there's nothing to return, and the caller of my function would do the same exact comparison. So how come this cannot be guaranteed to be a null pointer?

[Athar]

After all, I am comparing the pointer there to a value.

Not exactly, you're converting it to bool, which yields false if ptr has a null pointer value and true otherwise.
In the explicit comparison, you're explicitly comparing it to the null pointer value. Since 0 is a constant expression, the conversion to the null pointer value is guaranteed here.

But it's not guaranteed in the following case:

ptrdiff_t n=0;
if (ptr==reinterpret_cast<char*>(n))...

[Disch]

I'm having a hard time believing that, Athar.

I fail to see how a zero constant would ever be cast to a nonzero constant, regardless of what the types are.

The only way I can see that happening would be if the compiler was doing pointer math behind the scenes (casting between parent/child pointer types in a class hierarchy), but that wouldn't be the case here.

[Athar]

The point is that the null pointer value can be something other than zero.
There are few systems where this is the case, although none of them matter in every-day software development.
http://c-faq.com/null/machexamp.html

However, the internal representation shouldn't matter anyway in a valid C++ program.

[1vRz3TCk]

the standard wrote:

A value of integral type or enumeration type can be explicitly converted to a pointer.⁶⁴⁾ A pointer converted to an integer of sufficient size (if any such exists on the implementation) and back to the same pointer type will have its original value; mappings between pointers and integers are otherwise implementation-defined.
...
64)Converting an integral constant expression with value zero always yields a null pointer, but converting other expressions that happen to have value zero need not yield a null pointer.
page 75

[Disch]

Very interesting!

[Gaminic]

...really?