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*).

假设你有一个重载的函数(f)，它同时接受int和char*。在c++ 11之前，如果你想用空指针调用它，并且你使用了null(即值0)，那么你会调用int重载的指针:

void f(int);
void f(char*);

void g() 
{
  f(0); // Calls f(int).
  f(NULL); // Equals to f(0). Calls f(int).
}

这可能不是你想要的。c++ 11用nullptr解决了这个问题;现在你可以这样写:

void g()
{
  f(nullptr); //calls f(char*)
}

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.

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*).

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))。

从nullptr:一个类型安全和明确的空指针:

新的c++ 09 nullptr关键字指定了一个右值常量，用作通用空指针字面量，取代了有bug且弱类型的字面量0和臭名昭著的null宏。因此，Nullptr结束了30多年来的尴尬、歧义和错误。下面几节介绍nullptr功能，并展示它如何补救NULL和0的问题。

其他参考资料:

WikiBooks，带有示例代码。 Stack Overflow:在c++中，指针使用NULL还是0(零)? 模板 谷歌组:comp.lang.c++。有节制的编译器讨论