它是关键字，因为标准将这样指定它。;-)根据最新公开草案(n2914)

2.14.7指针字面量[lex.nullptr] pointer-literal: nullptr 指针字面值是关键字nullptr。它是std::nullptr_t类型的右值。

它很有用，因为它没有隐式地转换成一个整数值。

其他语言有保留词，它们是类型的实例。例如，Python:

>>> None = 5
  File "<stdin>", line 1
SyntaxError: assignment to None
>>> type(None)
<type 'NoneType'>

这实际上是一个相当接近的比较，因为None通常用于尚未初始化的东西，但与此同时，像None == 0这样的比较是假的。

另一方面，在普通C中，NULL == 0将返回真IIRC，因为NULL只是一个返回0的宏，这总是一个无效地址(AFAIK)。

让我首先给您一个简单的nullptr_t的实现

struct nullptr_t 
{
    void operator&() const = delete;  // Can't take address of nullptr

    template<class T>
    inline operator T*() const { return 0; }

    template<class C, class T>
    inline operator T C::*() const { return 0; }
};

nullptr_t nullptr;

nullptr是返回类型解析器习惯用法的一个微妙示例，它根据分配给实例的类型自动推导出正确类型的空指针。

int *ptr = nullptr;                // OK
void (C::*method_ptr)() = nullptr; // OK

如上所述，当nullptr被赋值给整数指针时，将创建模板化转换函数的int类型实例化。方法指针也是一样。 通过这种方式利用模板功能，我们实际上每次都创建了适当类型的空指针，这是一个新的类型赋值。 因为nullptr是一个值为0的整型字面值，你不能使用它的地址，这是我们通过删除&操作符实现的。

为什么我们首先需要nullptr ?

你可以看到传统的NULL有一些问题，如下:

1️⃣隐式转换

char *str = NULL; // Implicit conversion from void * to char *
int i = NULL;     // OK, but `i` is not pointer type

2️⃣函数调用歧义

void func(int) {}
void func(int*){}
void func(bool){}

func(NULL);     // Which one to call?

编译会产生以下错误:

error: call to 'func' is ambiguous
    func(NULL);
    ^~~~
note: candidate function void func(bool){}
                              ^
note: candidate function void func(int*){}
                              ^
note: candidate function void func(int){}
                              ^
1 error generated.
compiler exit status 1

3️⃣构造函数重载

struct String
{
    String(uint32_t)    {   /* size of string */    }
    String(const char*) {       /* string */        }
};

String s1( NULL );
String s2( 5 );

在这种情况下，需要显式强制转换(即String s((char*)0))。

它如何既是关键字又是类型的实例?

这并不奇怪。true和false都是关键字，作为字面量，它们有一个类型(bool)。Nullptr是一个std::nullptr_t类型的指针字面值，它是一个prvalue(不能使用&获取它的地址)。

4.10 about pointer conversion says that a prvalue of type std::nullptr_t is a null pointer constant, and that an integral null pointer constant can be converted to std::nullptr_t. The opposite direction is not allowed. This allows overloading a function for both pointers and integers, and passing nullptr to select the pointer version. Passing NULL or 0 would confusingly select the int version. A cast of nullptr_t to an integral type needs a reinterpret_cast, and has the same semantics as a cast of (void*)0 to an integral type (mapping implementation defined). A reinterpret_cast cannot convert nullptr_t to any pointer type. Rely on the implicit conversion if possible or use static_cast. The Standard requires that sizeof(nullptr_t) be sizeof(void*).

0 used to be the only integer value that could be used as a cast-free initializer for pointers: you can not initialize pointers with other integer values without a cast. You can consider 0 as a consexpr singleton syntactically similar to an integer literal. It can initiate any pointer or integer. But surprisingly, you'll find that it has no distinct type: it is an int. So how come 0 can initialize pointers and 1 cannot? A practical answer was we need a means of defining pointer null value and direct implicit conversion of int to a pointer is error-prone. Thus 0 became a real freak weirdo beast out of the prehistoric era. nullptr was proposed to be a real singleton constexpr representation of null value to initialize pointers. It can not be used to directly initialize integers and eliminates ambiguities involved with defining NULL in terms of 0. nullptr could be defined as a library using std syntax but semantically looked to be a missing core component. NULL is now deprecated in favor of nullptr, unless some library decides to define it as nullptr.

此外，你是否有另一个例子(除了维基百科的一个)，其中nullptr优于好旧的0?

是的。这也是在我们的生产代码中出现的一个(简化的)真实例子。它之所以突出，是因为gcc能够在交叉编译到具有不同寄存器宽度的平台时发出警告(仍然不确定为什么只有在从x86_64交叉编译到x86时，警告警告:从NULL转换为非指针类型'int'):

考虑以下代码(c++ 03):

#include <iostream>

struct B {};

struct A
{
    operator B*() {return 0;}
    operator bool() {return true;}
};

int main()
{
    A a;
    B* pb = 0;
    typedef void* null_ptr_t;
    null_ptr_t null = 0;

    std::cout << "(a == pb): " << (a == pb) << std::endl;
    std::cout << "(a == 0): " << (a == 0) << std::endl; // no warning
    std::cout << "(a == NULL): " << (a == NULL) << std::endl; // warns sometimes
    std::cout << "(a == null): " << (a == null) << std::endl;
}

它产生如下输出:

(a == pb): 1
(a == 0): 0
(a == NULL): 0
(a == null): 1