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

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.

其他回答

这是LLVM头文件。

// -*- C++ -*-
//===--------------------------- __nullptr --------------------------------===//
//
// Part of the LLVM Project, under the Apache License v2.0 with LLVM Exceptions.
// See https://llvm.org/LICENSE.txt for license information.
// SPDX-License-Identifier: Apache-2.0 WITH LLVM-exception
//
//===----------------------------------------------------------------------===//

#ifndef _LIBCPP_NULLPTR
#define _LIBCPP_NULLPTR

#include <__config>

#if !defined(_LIBCPP_HAS_NO_PRAGMA_SYSTEM_HEADER)
#pragma GCC system_header
#endif

#ifdef _LIBCPP_HAS_NO_NULLPTR

_LIBCPP_BEGIN_NAMESPACE_STD

struct _LIBCPP_TEMPLATE_VIS nullptr_t
{
    void* __lx;

    struct __nat {int __for_bool_;};

    _LIBCPP_INLINE_VISIBILITY _LIBCPP_CONSTEXPR nullptr_t() : __lx(0) {}
    _LIBCPP_INLINE_VISIBILITY _LIBCPP_CONSTEXPR nullptr_t(int __nat::*) : __lx(0) {}

    _LIBCPP_INLINE_VISIBILITY _LIBCPP_CONSTEXPR operator int __nat::*() const {return 0;}

    template <class _Tp>
        _LIBCPP_INLINE_VISIBILITY _LIBCPP_CONSTEXPR
        operator _Tp* () const {return 0;}

    template <class _Tp, class _Up>
        _LIBCPP_INLINE_VISIBILITY
        operator _Tp _Up::* () const {return 0;}

    friend _LIBCPP_INLINE_VISIBILITY _LIBCPP_CONSTEXPR bool operator==(nullptr_t, nullptr_t) {return true;}
    friend _LIBCPP_INLINE_VISIBILITY _LIBCPP_CONSTEXPR bool operator!=(nullptr_t, nullptr_t) {return false;}
};

inline _LIBCPP_INLINE_VISIBILITY _LIBCPP_CONSTEXPR nullptr_t __get_nullptr_t() {return nullptr_t(0);}

#define nullptr _VSTD::__get_nullptr_t()

_LIBCPP_END_NAMESPACE_STD

#else  // _LIBCPP_HAS_NO_NULLPTR

namespace std
{
    typedef decltype(nullptr) nullptr_t;
}

#endif  // _LIBCPP_HAS_NO_NULLPTR

#endif  // _LIBCPP_NULLPTR

(使用grep -r /usr/include/* '可以发现很多内容)

一个突出的东西是操作符* overload(返回0比分段故障友好得多……) 另一件事是,它看起来与存储地址根本不兼容。与抛出void*和将NULL结果作为哨兵值传递给普通指针的方式相比,这显然会减少“永远不要忘记,它可能是一个炸弹”的因素。

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)。

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*)
}