Understanding the interaction between bitfield and enumeration in S-function.

Question

Franck Yin 2018-6-15

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/405874-understanding-the-interaction-between-bitfield-and-enumeration-in-s-function

编辑： Franck Yin 2018-6-20

I have a manual code in which I have a structure with a bitfield with said enum's type:

typedef enum myEnum {
   MY_ENUM_VALUE_0 = 0,
   MY_ENUM_VALUE_1 = 1,
   MY_ENUM_VALUE_2 = 2,
   MY_ENUM_VALUE_3 = 3
} myEnum;
typedef struct myStructType {
   myEnum           enumVar  :2;
   unsigned char    filling  :6;
} myStructType;

I am using the legacy code tool to generate the S-function. This is my main function:

void mySFunctionMain (int* input, int* output)
{
   myStructType instance;
   instance.enumVar = *input;
   *output = instance.enumVar;
}

I observe the following input => output behaviour:

input = 0 => output = 0
input = 1 => output = 1
input = 2 => output = -2
input = 3 => output = -1

Can you please explain to me why this is happening ? If the bad conversion happens at the first attribution:

instance.enumVar = *input;

Then it means the enum on 2 bits is somehow considered signed and the value is changed there -even though it doesn't change anything about the fact that the enumVar bitfield data for e.g. input = 3 would now "11". If the bad conversion happens at the second attribution:

*output = instance.enumVar;

Then it would mean the data bits are correct but the enum would be incorrectly interpreted as signed on 2 bits.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Understanding the interaction between bitfield and enumeration in S-function.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

回答（0 个）

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

Understanding the interaction between bitfield and enumeration in S-function.

0 个评论 显示 -2更早的评论隐藏 -2更早的评论

回答（0 个）

另请参阅

类别

标签

产品

版本

Community Treasure Hunt

0 个评论
显示 -2更早的评论隐藏 -2更早的评论