C++ C的通用数组长度宏？
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Common array length macro for C?
提问by Matt Joiner
I've seen several macros for array length floating around:
我已经看到了几个数组长度的宏：
From this question:
从这个问题：
#define length(array) (sizeof(array)/sizeof(*(array)))#define ARRAY_LENGTH(array) (sizeof((array))/sizeof((array)[0]))#define SIZE(array, type) (sizeof(array) / (sizeof(type))
#define length(array) (sizeof(array)/sizeof(*(array)))#define ARRAY_LENGTH(array) (sizeof((array))/sizeof((array)[0]))#define SIZE(array, type) (sizeof(array) / (sizeof(type))
And Visual Studio's _countof:
和 Visual Studio 的_countof：
#define _countof(_Array) (sizeof(_Array) / sizeof(_Array[0]))
What I'd like to know is:
我想知道的是：
- What's the difference between those using
array[0]and*array? - Why should either be preferred?
- Do they differ in C++?
- 使用
array[0]和 的那些有什么区别*array？ - 为什么应该首选？
- 它们在 C++ 中有何不同？
回答by Michael Burr
Here's a better C version (from Google's Chromium project):
这是一个更好的 C 版本（来自 Google 的 Chromium 项目）：
#define COUNT_OF(x) ((sizeof(x)/sizeof(0[x])) / ((size_t)(!(sizeof(x) % sizeof(0[x])))))
It improves on the array[0]or *arrayversion by using 0[array], which is equivalent to array[0]on plain arrays, but will fail to compile if arrayhappens to be a C++ type that overloads operator[]().
它通过 using改进了array[0]or*array版本0[array]，这相当于array[0]在普通数组上，但如果array碰巧是重载 的 C++ 类型，将无法编译operator[]()。
The division causes a divide-by-zero operation (that should be caught at compile time since it's a compile-time constant expression) for many (but not all) situations where a pointer is passed as the arrayparameter.
在将指针作为array参数传递的许多（但不是全部）情况下，除法会导致除以零操作（应该在编译时捕获，因为它是编译时常量表达式）。
See Is there a standard function in C that would return the length of an array?for more details.
请参阅C 中是否有标准函数可以返回数组的长度？更多细节。
There's a better option for C++ code. See Compile time sizeof_array without using a macrofor details.
C++ 代码有更好的选择。有关详细信息，请参阅不使用宏的编译时 sizeof_array。
回答by Cheers and hth. - Alf
- What's the difference between those using array[0] and *array?
- Why should either be preferred?
- Do they differ in C++?
- 使用 array[0] 和 *array 的有什么区别？
- 为什么应该首选？
- 它们在 C++ 中有何不同？
(1) No difference in C. No difference for an actual raw array in C++.
(1) C 没有区别。C++ 中的实际原始数组没有区别。
(2) No technical grounds to prefer one or the other, but newbies might be confused by the pointer dereference.
(2) 没有技术依据来偏爱其中一个，但新手可能会被指针解引用混淆。
(3) In C++ you would normally not use the macro, because it's very unsafe. If you pass in a pointer instead of an actual raw array, code will compile but yield incorrect result. So in C++ you would/should instead use a function template, like …
(3) 在 C++ 中，您通常不会使用宏，因为它非常不安全。如果您传入一个指针而不是实际的原始数组，代码将编译但会产生错误的结果。因此，在 C++ 中，您将/应该使用函数模板，例如……
#include <stddef.h>
typedef ptrdiff_t Size;
template< class Type, Size n >
Size countOf( Type (&)[n] ) { return n; }
This only accepts actual raw array as argument.
这只接受实际的原始数组作为参数。
It's part of a triad of functions startOf, endOfand countOfthat it's very convenient to define so that they can be applied to both raw arrays and standard library containers. As far as I know this triad was first identified by Dietmar Kuehl. In C++0x startOfand endOfwill most probably be available as std::beginand std::end.
它的功能是黑社会的一部分startOf，endOf而且countOf，它是非常方便的定义，使他们能够同时适用于原数组和标准库容器。据我所知，这个黑社会最早是由迪特玛·库尔 (Dietmar Kuehl) 发现的。在 C++0x 中startOf，endOf很可能会以std::begin和 的形式提供std::end。
Cheers & hth.,
干杯 & hth.,
回答by SiegeX
1) Nothing, the value of an array is a pointer to it's first element. So *array == array[0]
2) Personal Preference
3) No
1) 没什么，数组的值是指向它的第一个元素的指针。所以 *array == array[0]
2) 个人偏好
3) 否
Note that this macro wont work if called inside a function where the array is passed as a parameter into the function. This is because the array object passed "decays" into a pointer rather than a deep copy.
请注意，如果在将数组作为参数传递给函数的函数内部调用该宏，则该宏将不起作用。这是因为数组对象将“衰减”传递给指针而不是深层副本。
