C++ 大小为 0 的数组
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Array with size 0
提问by zw324
Today I incidentally defined a two dimensional array with the size of one dimension being 0, however my compiler did not complain. I found the following which states that this is legal, at least in the case of gcc:
今天我顺便定义了一个二维数组，一维的大小为0，但是我的编译器没有抱怨。我发现以下内容表明这是合法的，至少在 gcc 的情况下：
However, I have two questions on this usage:
但是，我对这种用法有两个疑问：
First, is this considered as good programming practice? If so, then when should we use it in real world?
首先，这被认为是良好的编程习惯吗？如果是这样，那么我们什么时候应该在现实世界中使用它？
Second, the array I defined was two dimensional, with 0 size for one dimension. Is this the same as the one dimensional case? For example,
其次，我定义的数组是二维的，一维大小为 0。这和一维情况一样吗？例如，
int s[0]
int s[0][100]
int s[100][0]
Are they all the same in the memory and for the compiler?
它们在内存和编译器中都是一样的吗？
EDIT: Reply to Greg: The compiler I am using is gcc 4.4.5. My intention for this problem is not compiler-dependent, however if there are any compiler specific quirks that would be helpful too:)
编辑：回复 Greg：我使用的编译器是 gcc 4.4.5。我对这个问题的意图不依赖于编译器，但是如果有任何特定于编译器的怪癖也会有帮助:)
Thanks in advance!
提前致谢！
采纳答案by CB Bailey
In C++ it is illegal to declare an array of zero length. As such it is not normally considered a good practice as you are tying your code to a particular compiler extension. Many uses of dynamically sized arrays are better replaced with a container class such as std::vector.
在 C++ 中，声明长度为零的数组是非法的。因此，当您将代码绑定到特定的编译器扩展时，它通常不被认为是一种好的做法。动态大小数组的许多用途最好替换为容器类，例如std::vector.
ISO/IEC 14882:2003 8.3.4/1:
ISO/IEC 14882:2003 8.3.4/1：
If the constant-expression(5.19) is present, it shall be an integral constant expression and its value shall be greater than zero.
如果常量表达式(5.19) 存在，它应该是一个整数常量表达式并且它的值应该大于零。
However, you can dynamically allocate an array of zero length with new[].
但是，您可以使用 动态分配零长度的数组new[]。
ISO/IEC 14882:2003 5.3.4/6:
ISO/IEC 14882:2003 5.3.4/6：
The expression in a direct-new-declaratorshall have integral or enumeration type (3.9.1) with a non-negative value.
direct-new-declarator 中的表达式应具有非负值的整数或枚举类型 (3.9.1)。
回答by Node
I ran this program at ideone.com
我在ideone.com 上运行了这个程序
#include <iostream>
int main()
{
int a[0];
int b[0][100];
int c[100][0];
std::cout << "sizeof(a) = " << sizeof(a) << std::endl;
std::cout << "sizeof(b) = " << sizeof(b) << std::endl;
std::cout << "sizeof(c) = " << sizeof(c) << std::endl;
return 0;
}
It gave the size of all the variables as 0.
它将所有变量的大小设为 0。
sizeof(a) = 0
sizeof(b) = 0
sizeof(c) = 0
So in the above example, no memory is allocated for a, bor c.
所以在上面的例子中，没有为a,b或分配内存c。
回答by Kanopus
Compiling your example with gcc, all three of them have sizeof0, so I would assume that all of them are treated equally by the compiler.
用 gcc 编译你的例子，它们三个都有sizeof0，所以我假设所有这些都被编译器同等对待。
回答by Piotr Praszmo
Your link explains everything. They are used as last field in a struct when the length of struct is not known at compile time. If you try using them on stack or in a middle of other declarations you will end up overwriting next elements.
您的链接说明了一切。当编译时不知道结构体的长度时，它们用作结构体中的最后一个字段。如果您尝试在堆栈上或在其他声明的中间使用它们，您最终将覆盖下一个元素。
