C++快速判斷二進位制某位是1或0
阿新 • • 發佈:2018-12-26
其實原理很簡單,1BYTE = 8bit,bit是二進位制,根據8421編碼就很容易得出相對應的位,8bit中分為高4位和低4位,這樣可以從低位到高位 bit0-bit7 得到表示為:
0x01 0x02 0x04 0x08 0x10 0x20 0x40 0x80
得到的運算結果為:#include <iostream> using namespace std; typedef unsigned char UINT8; // 判斷某位是否為1 bool JudgeBitIsOne(UINT8 ucData, UINT8 ucFlag) { if (0x00 == ucData) return false; else if (ucData == ucFlag) return true; } int _tmain(int argc, _TCHAR* argv[]) { UINT8 ucTemp = 0xF2; // 1111 0010 // bit0-bit7 0x01 0x02 0x04 0x08 0x10 0x20 0x40 0x80 cout << JudgeBitIsOne((ucTemp & 0x01), 0x01) << " "; cout << JudgeBitIsOne((ucTemp & 0x02), 0x02) << " "; cout << JudgeBitIsOne((ucTemp & 0x04), 0x04) << " "; cout << JudgeBitIsOne((ucTemp & 0x08), 0x08) << " "; cout << JudgeBitIsOne((ucTemp & 0x10), 0x10) << " "; cout << JudgeBitIsOne((ucTemp & 0x20), 0x20) << " "; cout << JudgeBitIsOne((ucTemp & 0x40), 0x40) << " "; cout << JudgeBitIsOne((ucTemp & 0x80), 0x80) << " " << endl; return 0; }