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.

Nullptr不能赋值给整型类型，比如int型，只能赋值给指针类型;内置指针类型，如int *ptr或智能指针，如std::shared_ptr<T>

我相信这是一个重要的区别，因为NULL仍然可以被赋给整型和指针，因为NULL是一个扩展到0的宏，可以作为int的初始值以及指针。

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))。

当一个函数可以接收指向多个类型的指针时，用NULL来调用它是不明确的。这种工作方式现在是非常hack的接受一个int并假设它是NULL。

template <class T>
class ptr {
    T* p_;
    public:
        ptr(T* p) : p_(p) {}

        template <class U>
        ptr(U* u) : p_(dynamic_cast<T*>(u)) { }

        // Without this ptr<T> p(NULL) would be ambiguous
        ptr(int null) : p_(NULL)  { assert(null == NULL); }
};

在c++ 11中，你可以重载nullptr_t，这样ptr<T> p(42);将是编译时错误，而不是运行时断言。

ptr(std::nullptr_t) : p_(nullptr)  {  }

NULL need not to be 0. As long you use always NULL and never 0, NULL can be any value. Asuming you programme a von Neuman Microcontroller with flat memory, that has its interrupt vektors at 0. If NULL is 0 and something writes at a NULL Pointer the Microcontroller crashes. If NULL is lets say 1024 and at 1024 there is a reserved variable, the write won't crash it, and you can detect NULL Pointer assignments from inside the programme. This is Pointless on PCs, but for space probes, military or medical equipment it is important not to crash.

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*)
}