从c ++ struct字段获取单个字段的大小 [英] Getting the size of an indiviual field from a c++ struct field

问题描述

简短版本是：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?

为了阐明，字段的示例我在说：

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.

要获得整个Test对象的大小很容易，只是说

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

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

我想知道如何获得个别字段，说Test :: field4。上面的例子对一个普通的结构成员不起作用。有任何想法吗？或者有人知道一个原因，为什么它不能工作？我相信sizeof不会有帮助，因为它只返回字节大小，但如果有人知道，否则我都耳朵。

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！ >

Thanks!

推荐答案

您可以计算运行时的大小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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