问题描述
我有一个设计,要求值包含在32位字内的某些位。示例是位10-15必须保持值9,其余位都为0.因此为了简单/可读性,我创建了一个结构,其中包含所要求的分解版本。
I have a design that requires values to be contained at certain bits inside of a 32 bit word. Example being bits 10-15 must hold value 9, with the remaining bits all being 0. So for simplicity/readability I created a struct that contains a broken down version of what is asked.
struct {
int part1 : 10;
int part2 : 6;
int part3 : 16;
} word;
然后我可以设置 part2
word.part1 = 0;
word.part2 = 9;
word.part3 = 0;
现在我想使用该结构,并将其转换为单个32位整数。我有它通过强制转换编译,但它似乎不是一个非常优雅或安全的方式转换数据。
I now want to take that struct, and convert it into a single 32 bit integer. I do have it compiling by forcing the casting, but it does not seem like a very elegant or secure way of converting the data.
int x = *reinterpret_cast<int*>(&word);
如果我试图把它转换为一个普通的 reinterpret_cast< int> word)
我收到以下错误:
If I try to cast it just as a normal reinterpret_cast<int>(word)
I get the following error:
invalid cast from type 'ClassName::<anonymous struct>' to type 'int'
必须有一个更好的方法,没有想出来。提前感谢!
There must be a better way of doing this, I just can not figure it out. Thanks in advance!
注意:必须使用c ++风格的转换,因为标准和其他...
Note: Must be done in c++ style casting, because of standards and whatnot... eye roll
推荐答案
尝试 reinterpret_cast< int>(word)
转换运算符在用户定义的结构类型和 int
之间定义。
The attempt reinterpret_cast<int>(word)
does not work because there is no conversion operator defined between your user-defined struct type and int
.
您可以为结构添加转换运算符,或者最好使用IMHO命名函数进行转换,例如:
You could add a conversion operator to your struct, or preferably IMHO a named function to do the conversion, e.g.:
struct {
uint32_t part1 : 10;
uint32_t part2 : 6;
uint32_t part3 : 16;
uint32_t get_all_parts() const
{
return (part1 << 22) + (part2 << 16) + part3;
}
} word;
注意,我使用无符号整数,因为他们在左移时有明确定义的行为。
Note, I used unsigned ints as they have well-defined behaviour on left shifting.
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