现在c++ 11有了许多新特性。一个有趣而令人困惑的(至少对我来说)是新的nullptr。

不需要讨厌的宏NULL了。

int* x = nullptr;
myclass* obj = nullptr;

不过,我还是不明白nullptr是如何工作的。例如,维基百科的一篇文章说:

c++ 11通过引入一个新的关键字作为区分空指针常量nullptr来纠正这一点。它的类型为nullptr_t,可隐式转换,可与任何指针类型或指针到成员类型相比较。它不能隐式转换,也不能与整型相比,bool类型除外。

它如何既是关键字又是类型的实例?

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?


当前回答

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

其他回答

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*).

为什么在c++ 11中使用nullptr ?是什么?为什么NULL是不充分的?

c++专家Alex Allain在这里说得很好(我用粗体加了重点):

...imagine you have the following two function declarations: void func(int n); void func(char *s); func( NULL ); // guess which function gets called? Although it looks like the second function will be called--you are, after all, passing in what seems to be a pointer--it's really the first function that will be called! The trouble is that because NULL is 0, and 0 is an integer, the first version of func will be called instead. This is the kind of thing that, yes, doesn't happen all the time, but when it does happen, is extremely frustrating and confusing. If you didn't know the details of what is going on, it might well look like a compiler bug. A language feature that looks like a compiler bug is, well, not something you want. Enter nullptr. In C++11, nullptr is a new keyword that can (and should!) be used to represent NULL pointers; in other words, wherever you were writing NULL before, you should use nullptr instead. It's no more clear to you, the programmer, (everyone knows what NULL means), but it's more explicit to the compiler, which will no longer see 0s everywhere being used to have special meaning when used as a pointer.

Allain在文章结尾写道:

不管这些——c++ 11的经验法则是,只要在过去使用NULL,就开始使用nullptr。

(我的话):

最后,不要忘记nullptr是一个对象——一个类。它可以在任何之前使用NULL的地方使用,但如果你因为某种原因需要它的类型,它的类型可以用decltype(nullptr)提取,或者直接描述为std::nullptr_t,这只是一个decltype(nullptr)的类型定义,如下所示:

定义在头<cstddef>:

See:

https://en.cppreference.com/w/cpp/types/nullptr_t 和https://en.cppreference.com/w/cpp/header/cstddef

namespace std
{
typedef decltype(nullptr) nullptr_t; // (since C++11)
// OR (same thing, but using the C++ keyword `using` instead of the C and C++ 
// keyword `typedef`):
using nullptr_t = decltype(nullptr); // (since C++11)
} // namespace std

引用:

c++ 11中更好的类型- nullptr,枚举类(强类型枚举)和cstdint https://en.cppreference.com/w/cpp/language/decltype https://en.cppreference.com/w/cpp/types/nullptr_t https://en.cppreference.com/w/cpp/header/cstddef https://en.cppreference.com/w/cpp/keyword/using https://en.cppreference.com/w/cpp/keyword/typedef

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