从 C++ struct 字段获取单个字段的大小

Getting the size of an indiviual field from a c++ struct field(从 C++ struct 字段获取单个字段的大小)

问题描述

简短版本是:我如何了解 c++ 字段的单个字段的大小(以位为单位)?

The short version is: How do I learn the size (in bits) of an individual field of a c++ field?

为了澄清，我正在谈论的领域的一个例子:

To clarify, an example of the field I am talking about:

struct Test {
    unsigned field1 : 4;  // takes up 4 bits
    unsigned field2 : 8;  // 8 bits
    unsigned field3 : 1;  // 1 bit
    unsigned field4 : 3;  // 3 bits
    unsigned field5 : 16; // 16 more to make it a 32 bit struct

    int normal_member; // normal struct variable member, 4 bytes on my system
};

Test t;
t.field1 = 1;
t.field2 = 5;
// etc.

获取整个Test对象的大小很简单，我们就说

To get the size of the entire Test object is easy, we just say

sizeof(Test); // returns 8, for 8 bytes total size

我们可以通过获取一个普通的struct成员

We can get a normal struct member through

sizeof(((Test*)0)->normal_member); // returns 4 (on my system)

我想知道如何获取单个字段的大小，例如 Test::field4.普通结构成员的上述示例不起作用.有任何想法吗?或者有人知道它不能工作的原因吗?我相当相信 sizeof 不会有帮助，因为它只以字节为单位返回大小，但如果有人知道否则我会全神贯注.

I would like to know how to get the size of an individual field, say Test::field4. The above example for a normal struct member does not work. Any ideas? Or does someone know a reason why it cannot work? I am fairly convinced that sizeof will not be of help since it only returns size in bytes, but if anyone knows otherwise I'm all ears.

谢谢！

推荐答案

可以在运行时计算大小，fwiw，例如:

You can calculate the size at run time, fwiw, e.g.:

//instantiate
Test t;
//fill all bits in the field
t.field1 = ~0;
//extract to unsigned integer
unsigned int i = t.field1;
... TODO use contents of i to calculate the bit-width of the field ...

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