如何使用位字段为结构赋值？

Question

我有一个带有位字段的结构(总共32位宽度),我有一个32位变量.当我尝试将变量值分配给我的结构时,我收到一个错误:

错误:从'uint32_t {aka unsigned int}'转换为请求的非标量类型'main():: CPUID'.

struct CPUIDregs
    {
       uint32_t EAXBuf;
    };
CPUIDregs CPUIDregsoutput;   


int main () {

 struct CPUID          
    {
          uint32_t   Stepping         : 4;         
          uint32_t   Model            : 4;        
          uint32_t   FamilyID         : 4;        
          uint32_t   Type             : 2;        
          uint32_t   Reserved1        : 2;         
          uint32_t   ExtendedModel    : 4;         
          uint32_t   ExtendedFamilyID : 8;          
          uint32_t   Reserved2        : 4;          
    };

    CPUID CPUIDoutput = CPUIDregsoutput.EAXBuf;

你知道如何以最短的方式做到这一点吗？谢谢

PS当然我在实际代码中有更合适的EAX值,但我想这不会影响到这里.

Answer 1

您永远不应该依赖编译器如何在内存中布局您的结构.有一些方法可以通过单一作业完成您想要的任务,但我既不会推荐也不会告诉您.

完成任务的最佳方式如下:

static inline void to_id(struct CPUid *id, uint32_t value)
{
    id->Stepping         = value & 0xf;
    id->Model            = value >> 4 & 0xf;
    id->FamilyID         = value >> 8 & 0xf;
    id->Type             = value >> 12 & 0x3;
    id->Reserved1        = value >> 14 & 0x3;
    id->ExtendedModel    = value >> 16 & 0xf;
    id->ExtendedFamilyID = value >> 20 & 0xff;
    id->Reserved2        = value >> 28 & 0xf;
}

与此相反

static inline uint32_t from_id(struct CPUid *id)
{
    return id->Stepping
         | id->Model << 4
         | id->FamilyID << 8
         | id->Type << 12
         | id->Reserved1 << 14
         | id->ExtendedModel << 16
         | id->ExtendedFamilyID << 20
         | id->Reserved2 << 28;
}

Answer 2

使用联合。

union foo {
    struct {
        uint8_t a : 4;
        uint8_t b : 4;
        uint8_t c : 4;
        uint8_t d : 4;
        uint16_t e;
    };
    uint32_t allfields;
};

int main(void) {
    union foo a;

    a.allfields = 0;
    a.b = 3;

    return 0;
}