什么是通用的C / C ++宏来确定结构成员的大小？ [英] What is a common C/C++ macro to determine the size of a structure member?

问题描述

在C / C ++中，如何确定结构的成员变量的大小，而无需定义该结构类型的虚拟变量？这里有一个例子，说明如何做错了，但显示的意图：

In C/C++, how do I determine the size of the member variable to a structure without needing to define a dummy variable of that structure type? Here's an example of how to do it wrong, but shows the intent:

typedef struct myStruct {
  int x[10];
  int y;
} myStruct_t;

const size_t sizeof_MyStruct_x = sizeof(myStruct_t.x);  // error

为了参考，这应该是如何找到'x'定义一个虚拟变量：

For reference, this should be how to find the size of 'x' if you first define a dummy variable:

myStruct_t dummyStructVar;

const size_t sizeof_MyStruct_x = sizeof(dummyStructVar.x);

但是，我希望避免创建一个虚拟变量， X'。我认为有一个聪明的方式重写0作为一个myStruct_t帮助找到成员变量'x'的大小，但它已经足够长，我已经忘记了细节，似乎不能得到一个好的谷歌搜索这个。你知道吗？

However, I'm hoping to avoid having to create a dummy variable just to get the size of 'x'. I think there's a clever way to recast 0 as a myStruct_t to help find the size of member variable 'x', but it's been long enough that I've forgotten the details, and can't seem to get a good Google search on this. Do you know?

谢谢！

推荐答案

是标签说的），你的虚拟变量代码可以替换为：

In C++ (which is what the tags say), your "dummy variable" code can be replaced with:

sizeof myStruct_t().x;

不会创建myStruct_t对象：编译器只能处理sizeof操作数的静态类型，

No myStruct_t object will be created: the compiler only works out the static type of sizeof's operand, it doesn't execute the expression.

这在C中工作，在C ++中更好，因为它也适用于没有可访问的无参数构造函数的类：

This works in C, and in C++ is better because it also works for classes without an accessible no-args constructor:

sizeof ((myStruct_t *)0)->x

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