Question

I found a bug in my code where I compared the pointer with ‘\0’.

Wondering why the compiler didn’t warn me about this bug I tried the following.

#include <cassert>

struct Foo
{
    char bar[5];
};

int main()
{
    Foo f;
    Foo* p = &f;
    p->bar[0] = '\0';
    assert(p->bar == '\0');    // #1. I forgot [] Now, comparing pointer with NULL and fails.
    assert(p->bar == 'A');     // #2. error: ISO C++ forbids comparison between pointer and integer
    assert(p->bar[0] == '\0'); // #3. What I intended, PASSES
    return 0;
}

What is special about ‘\0’ which makes #1 legal and #2 illegal?

Please add a reference or quotation to your answer.

Answer 1

What makes it legal and well defined is the fact that '\0' is a null pointer constant so it can be converted to any pointer type to make a null pointer value.

ISO/IEC 14882:2011 4.10 [conv.ptr] / 1:

A null pointer constant is an integral constant expression prvalue of integer type that evaluates to 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 and is distinguishable from every other value of object pointer or function pointer type. Such a conversion is called a null pointer conversion.

'\0' meets the requirements of “integral constant expression prvalue of integer type that evaluates to zero” because char is an integer type and \0 has the value zero.

Other integers can only be explicitly converted to a pointer type via a reinterpret_cast and the result is only meaningful if the integer was the result of converting a valid pointer to an integer type of sufficient size.