【高手】小字节序转为大字节序函数

【高手求助】小字节序转为大字节序函数
昨天的一道笔试题，编写如下结构的小字节序转大字节序的转换函数：void ConvertLToB(T_Sample*ptIn),结构的定义如下

C/C++ code

#pragma pack(4)typedef struct tagT_Sample{BYTE ucA   :3;BYTE ucB   :3;BYTE ucC   :3;BYTE ucD   :3;BYTE ucE   :3;DWORD dwF   :24;}T_Sample;

自己对大小端也有点了解，但是也只是限于判断和求个输出什么的，哎，怪自己学艺不精，求高手来解答！

[解决办法]
#pragma pack(4)
typedef struct tagT_Sample{
////1字节//////
BYTE ucA :3;
BYTE ucB :3;
//////////////

////1字节/////
BYTE ucC :3;
BYTE ucD :3;
//////////////

///1字节//////
BYTE ucE :3;
//////////////

//填充1字节///
/////////////

/////4字节////
DWORD dwF :24;
/////////////

}T_Sample;


就这么8个字节, SB面试官是打算让你把整个结构体的字节颠倒一下还是把DWORD字节的字节序颠倒一下(唯一的多字节),还是把每个域里的位颠倒一下?
[解决办法]
出题的就是个半桶水！

按照标准sizeof(T_Sample)都是不确定的！

对于位字段来说，几乎所有的描述都是不确定的implementation-defined或者unspecified.

A bit-field may have type int, unsigned int, or signed int. Whether the high-order bit position of a “plain” int bit-field is treated as a sign bit is implementation-defined. A bit-field is interpreted as an integral type consisting of the specified number of bits.

An implementation may allocate any addressable storage unit large enough to hold a bit-field. If enough space remains, a bit-field that immediately follows another bit-field in a structure shall be packed into adjacent bits of the same unit. If insufficient space remains, whether a bit-field that does not fit is put into the next unit or overlaps adjacent units is implementation-defined. The order of allocation of bit-fields within a unit (high-order to low-order or low-order to high-order) is implementation-defined. The alignment of the addressable storage unit is unspecified.

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