Strange behavior with null pointers

Question

I have stumbled upon the following problem, if I compile (using Visual Studio 2010)

int main()
{
    double* d_ptr = int(0);
    if(d_ptr == nullptr)
         cout << "I am a nullptr" << endl;

    return 0;
}

I get the result "I am a nullptr"

This also works if I substitute the pointer assignment with:

double* ptr = (int) 0;

I would have expected both to fail since they both explicitly cast to integer which has not a pointer type. Could someone help me in understanding what is going on?

EDIT: Tried again with g++ and worked. Do not know what I did wrong the first time. Now I am only still troubled why it works in the first place.

[Works for me](http://coliru.stacked-crooked.com/a/8a4516f5d3e9750e) with g++. — NathanOliver, Sep 08 '16 at 15:51
@NathanOliver clang says ["cannot initialize a variable of type 'double *' with an rvalue of type 'int'"](http://coliru.stacked-crooked.com/a/ad432ad2fb0412cb), though. — jaggedSpire, Sep 08 '16 at 15:54
Please check out [Do you use NULL or 0 (zero) for pointers in C++?](http://stackoverflow.com/questions/176989/do-you-use-null-or-0-zero-for-pointers-in-c) — Bo Persson, Sep 08 '16 at 15:57

Martin Bonner supports Monica · Accepted Answer · 2016-09-08T16:19:26.677

4

You have wandered into one of the murkier areas of C++ (and C). Quoting from n4296 (which is the latest freely available draft of C++14):

4.10 Pointer conversions [conv.ptr]

1 A null pointer constant is an integer literal (2.13.2) with value zero or a prvalue of type std::nullptr_t. A null pointer constant can be converted to a pointer type; the result is the null pointer value of that type

I am slightly surprised that (int)0 is accepted as "an integer literal" - in fact, I think that is a bug in the compiler (but it is very much a corner case).

Ah-ha! I have just remembered: Older versions of C and C++ didn't say "integer literal", they said "integer constant expression". int(0) is an integer constant expression, so it is allowed on older compilers.

This means your problem with g++ could depend on whether you specified C++14 or an earlier version.

edited Sep 08 '16 at 16:19

answered Sep 08 '16 at 16:15

Martin Bonner supports Monica

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I believe enforcing `(int) 0` to fail would possibly break way too match existing code. – Slava Sep 08 '16 at 16:18
@Slava: You think so? Most people write `NULL` or `0` - not `(int)0`. The cast is pointless as `0` is already of type int. – Martin Bonner supports Monica Sep 08 '16 at 16:20
@MartinBonner (int)0 or other expressions used to be legal in old C++ standard. Restricting it may break working code anyway. – Slava Sep 08 '16 at 16:54
@DieterLücking: I don't *think* it should. That looks like a bug to me - but I could easily be wrong! – Martin Bonner supports Monica Sep 08 '16 at 18:49

score 0 · Answer 2 · answered Sep 08 '16 at 15:59

0

By assigning the value of 0, to the pointer, you are setting it to null, so it is a nullptr. To make the pointer have the value of 0, do this:

double * dbl = new double{0};

answered Sep 08 '16 at 15:59

Arnav Borborah

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Strange behavior with null pointers

2 Answers2