Is C NULL equal to C++11 nullptr

In C++11 and beyond, a pointer that is ==NULL will also ==nullptr and vice versa.

Uses of NULL other than comparing with a pointer (like using it to represent the nul byte at the end of a string) won’t work with nullptr.

In some cases, NULL is #define NULL 0, as the integer constant 0 is special-cased in C and C++ when you compare it with pointers. This non-type type information causes some problems in both C and C++, so in C++ they decided to create a special type and value that does the same thing in the “proper” use cases, and reliably fails to compile in most of the “improper” use cases.

Insofar as your C++ implementation is compatible with the C implementation you are interoping with (very rare for this not to be true), everything should work.

To be very clear, if ptr is any kind of pointer, then the following expressions are equivalent in C++:

ptr == nullptr
ptr == NULL
ptr == 0

As are the following:

ptr = nullptr
ptr = NULL
ptr = 0

and if X is some type, so are the following statements:

X* ptr = nullptr;
X* ptr = NULL;
X* ptr = 0;

nullptr differs when you pass it to a template function that deduces type (NULL or 0 become an int unless passed to an argument expecting a pointer, while nullptr remains a nullptr_t), and when used in some contexts where nullptr won’t compile (like char c = NULL;) (note, not char* c=NULL;)

Finally, literally:

NULL == nullptr

is true.

The NULL constant gets promoted to a pointer type, and as a pointer it is a null pointer, which then compares equal to nullptr.

Despite all this, it isn’t always true that:




do the same thing.

void bar(int) { std::cout << "int\n"; }
void bar(void*) { std::cout << "void*\n"; }
template<class T>
void foo(T t) { bar(t); }

this prints int for NULL and void* for nullptr.

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