推 LPH66:你引的標準那段話就是在說碰到剩下的不夠時要另開還是硬接 12/26 12:19
→ LPH66:是 implementation-defined 隨實作決定的 12/26 12:19
→ LPH66:那你的編譯器就只是選擇另開這個選項而已 12/26 12:20
→ illjay0404:了解!!所以空間的分配是由編譯器決定! 12/27 08:50
借標題詢問...
最近剛好也有碰到類似的問題 ,看完各位大大解釋後
又順便再google一下 ,發現衍伸性的問題
我使用平台是bcb5
#pragma pack(push,1)
struct SBitField {
unsigned char Seconds:5;
unsigned char Minutes:6;
unsigned char Hours :5;
};
#pragma pack(pop)
sizeof(SBitField) 結果是 3
又再google後發現...
C99 6.7.2.1 Structure and union specifier:
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.
然後自己又試了一下....
#pragma pack(push,1)
struct SBitField {
unsigned short Seconds:5;
unsigned short Minutes:6;
unsigned short Hours :5;
};
#pragma pack(pop)
sizeof(SBitField) 結果結果就變成2了
我想問的是~這感覺是不是像alignment ?
當編譯器配置時發現Char Minutes 類別空間不夠擺入 char Seconds
所以再配置另一個Char類別空間 ,擺放給char Seconds
然後char Hours也是 ,所以sizeof後是3 ?
short是直接大類別空間擺放bit-field所以是2 ?
請各問大大解除小弟疑疑惑 Orz...
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※ 編輯: illjay0404 來自: 59.124.107.139 (12/26 10:58)