为什么空指针在C和C ++不同的定义？ [英] Why are NULL pointers defined differently in C and C++?
问题描述
在C， NULL 定义为（无效*）0 ，而在C ++中是 0 。为什么会这样呢？
在C我能理解，如果 NULL 不强制转换为（无效*）然后编译器可能/可能不会产生警告。除此之外，还有什么原因？
In C, NULL is defined as (void *)0 whereas in C++ it is 0. Why is it so?
In C I can understand that if NULL is not typecast to (void *) then compilers may/may not generate warning. Other than this, is there any reason?
推荐答案
在C ++中，空指针是由ISO规范（宗; 4.10 / 1）定义为
In C++, the null pointer is defined by the ISO specification (§4.10/1) as
一个空指针常量是一个整型常量前pression（5.19）计算结果为零整数类型的右值。
A null pointer constant is an integral constant expression (5.19) rvalue of integer type that evaluates to zero.
这就是为什么在C ++中，你可以写
This is why in C++ you can write
int* ptr = 0;
在C，这个规则是相似的，但有点不同（宗; 6.3.2.3/3）：
In C, this rule is similar, but is a bit different (§6.3.2.3/3):
值为0，或者这样的前pression投整型常量前pression键入
无效* ，被称为一个空指针constant.55）如果空指针常量将被转换为
指针类型，所得到的指针，称为空指针，被保证比较不等
的指针的任何对象或功能。
An integer constant expression with the value 0, or such an expression cast to type
void *, is called a null pointer constant.55) If a null pointer constant is converted to a pointer type, the resulting pointer, called a null pointer, is guaranteed to compare unequal to a pointer to any object or function.
因此，无论
int* ptr = 0;
和
int* ptr = (void *)0
是合法的。不过，我的猜测是，无效* 投在这里，使之类的语句
are legal. However, my guess is that the void* cast is here so that statements like
int x = NULL;
产生在大多数系统上编译器警告。在C ++中，这不会是合法的，因为你不能隐式转换无效* 隐另一个指针类型不进行强制转换。例如，这是非法的：
produce a compiler warning on most systems. In C++, this wouldn't be legal because you can't implicitly convert a void* to another pointer type implicitly without a cast. For example, this is illegal:
int* ptr = (void*)0; // Legal C, illegal C++
然而，这将导致问题，因为code
However, this leads to issues because the code
int x = NULL;
是合法的C ++。因为这和随之而来的混乱（和另一种情况，后来所示），C ++ 11现在有一个关键字的 nullptr 重新presenting一个空指针：
is legal C++. Because of this and the ensuing confusion (and another case, shown later), C++11 now has a keyword nullptr representing a null pointer:
int* ptr = nullptr;
此没有任何的上述问题。
This doesn't have any of the above problems.
nullptr 超过0的另一个优点是，它起着与C ++类型系统更好。例如，假设我有这两个功能：
The other advantage of nullptr over 0 is that it plays better with the C++ type system. For example, suppose I have these two functions:
void DoSomething(int x);
void DoSomething(char* x);
如果我称之为
DoSomething(NULL);
这是等同于
DoSomething(0);
它调用 DoSomething的（INT）而不是预期的 DoSomething的（字符*）。然而，随着 nullptr ，我可以写
which calls DoSomething(int) instead of the expected DoSomething(char*). However, with nullptr, I could write
DoSomething(nullptr);
和它将调用 DoSomething的（字符*）功能如预期。
And it will call the DoSomething(char*) function as expected.
同样的，假设我有一个矢量＆lt;对象* GT; 并要设置的每个元素是一个空指针。使用的std ::填写算法，我可能会尝试写
Similarly, suppose that I have a vector<Object*> and want to set each element to be a null pointer. Using the std::fill algorithm, I might try writing
std::fill(v.begin(), v.end(), NULL);
不过，这并不能编译，因为模板系统治疗 NULL 为 INT ，而不是一个指针。为了解决这个问题，我会写
However, this doesn't compile, because the template system treats NULL as an int and not a pointer. To fix this, I would have to write
std::fill(v.begin(), v.end(), (Object*)NULL);
这是丑陋的，有点违背了模板系统的目的。为了解决这个问题，我可以使用 nullptr ：
This is ugly and somewhat defeats the purpose of the template system. To fix this, I can use nullptr:
std::fill(v.begin(), v.end(), nullptr);
和自 nullptr 是已知有对应于空指针（具体而言，的std :: nullptr_t A型），这将正确编译。
And since nullptr is known to have a type corresponding to a null pointer (specifically, std::nullptr_t), this will compile correctly.
希望这有助于！
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