如何设置、清除和切换一点?


当前回答

使用标准C++库:std::bitset<N>。

或者Boost版本:Boost::dynamic_bitset。

没有必要自己动手:

#include <bitset>
#include <iostream>

int main()
{
    std::bitset<5> x;

    x[1] = 1;
    x[2] = 0;
    // Note x[0-4]  valid

    std::cout << x << std::endl;
}

[Alpha:] > ./a.out
00010

与标准库编译时大小的位集相比,Boost版本允许运行时大小的比特集。

其他回答

以下是C中的一个例程,用于执行基本的逐位操作:

#define INT_BIT (unsigned int) (sizeof(unsigned int) * 8U) //number of bits in unsigned int

int main(void)
{
    
    unsigned int k = 5; //k is the bit position; here it is the 5th bit from the LSb (0th bit)
    
    unsigned int regA = 0x00007C7C; //we perform bitwise operations on regA
    
    regA |= (1U << k);    //Set kth bit
    
    regA &= ~(1U << k);   //Clear kth bit
    
    regA ^= (1U << k);    //Toggle kth bit
    
    regA = (regA << k) | regA >> (INT_BIT - k); //Rotate left by k bits
    
    regA = (regA >> k) | regA << (INT_BIT - k); //Rotate right by k bits

    return 0;   
}

比特场方法在嵌入式领域还有其他优势。您可以定义直接映射到特定硬件寄存器中的位的结构。

struct HwRegister {
    unsigned int errorFlag:1;  // one-bit flag field
    unsigned int Mode:3;       // three-bit mode field
    unsigned int StatusCode:4;  // four-bit status code
};

struct HwRegister CR3342_AReg;

您需要注意位打包顺序-我认为它首先是MSB,但这可能取决于实现。此外,验证编译器处理程序字段如何跨越字节边界。

然后,您可以像以前一样读取、写入和测试各个值。

如果你在做很多无聊的事情,你可能会想使用口罩,这会让整个事情变得更快。以下函数速度非常快,而且仍然灵活(它们允许在任何大小的位图中进行位旋转)。

const unsigned char TQuickByteMask[8] =
{
   0x01, 0x02, 0x04, 0x08,
   0x10, 0x20, 0x40, 0x80,
};


/** Set bit in any sized bit mask.
 *
 * @return    none
 *
 * @param     bit    - Bit number.
 * @param     bitmap - Pointer to bitmap.
 */
void TSetBit( short bit, unsigned char *bitmap)
{
    short n, x;

    x = bit / 8;        // Index to byte.
    n = bit % 8;        // Specific bit in byte.

    bitmap[x] |= TQuickByteMask[n];        // Set bit.
}


/** Reset bit in any sized mask.
 *
 * @return  None
 *
 * @param   bit    - Bit number.
 * @param   bitmap - Pointer to bitmap.
 */
void TResetBit( short bit, unsigned char *bitmap)
{
    short n, x;

    x = bit / 8;        // Index to byte.
    n = bit % 8;        // Specific bit in byte.

    bitmap[x] &= (~TQuickByteMask[n]);    // Reset bit.
}


/** Toggle bit in any sized bit mask.
 *
 * @return   none
 *
 * @param   bit    - Bit number.
 * @param   bitmap - Pointer to bitmap.
 */
void TToggleBit( short bit, unsigned char *bitmap)
{
    short n, x;

    x = bit / 8;        // Index to byte.
    n = bit % 8;        // Specific bit in byte.

    bitmap[x] ^= TQuickByteMask[n];        // Toggle bit.
}


/** Checks specified bit.
 *
 * @return  1 if bit set else 0.
 *
 * @param   bit    - Bit number.
 * @param   bitmap - Pointer to bitmap.
 */
short TIsBitSet( short bit, const unsigned char *bitmap)
{
    short n, x;

    x = bit / 8;    // Index to byte.
    n = bit % 8;    // Specific bit in byte.

    // Test bit (logigal AND).
    if (bitmap[x] & TQuickByteMask[n])
        return 1;

    return 0;
}


/** Checks specified bit.
 *
 * @return  1 if bit reset else 0.
 *
 * @param   bit    - Bit number.
 * @param   bitmap - Pointer to bitmap.
 */
short TIsBitReset( short bit, const unsigned char *bitmap)
{
    return TIsBitSet(bit, bitmap) ^ 1;
}


/** Count number of bits set in a bitmap.
 *
 * @return   Number of bits set.
 *
 * @param    bitmap - Pointer to bitmap.
 * @param    size   - Bitmap size (in bits).
 *
 * @note    Not very efficient in terms of execution speed. If you are doing
 *        some computationally intense stuff you may need a more complex
 *        implementation which would be faster (especially for big bitmaps).
 *        See (http://graphics.stanford.edu/~seander/bithacks.html).
 */
int TCountBits( const unsigned char *bitmap, int size)
{
    int i, count = 0;

    for (i=0; i<size; i++)
        if (TIsBitSet(i, bitmap))
            count++;

    return count;
}

注意,要在16位整数中设置位“n”,请执行以下操作:

TSetBit( n, &my_int);

这取决于你确保比特数在你通过的位图范围内。请注意,对于字节、字、dword、qword等的小端处理器,它们在内存中正确地相互映射(小端处理器比大端处理器“更好”的主要原因,啊,我感觉到一场火焰大战即将到来…)。

将第n位设置为x(位值),不使用-1

有时,当您不确定-1或类似的结果时,您可能希望在不使用-1的情况下设置第n位:

number = (((number | (1 << n)) ^ (1 << n))) | (x << n);

解释:((number |(1<<n)将第n位设置为1(其中|表示逐位OR),然后使用(…)^(1<<n)将将第n个位设置为0,最后通过(…)|x<<n,将第n比特设置为0(位值)x。

这也适用于戈朗。

如果您想在Linux内核中使用C编程执行所有操作,那么我建议使用Linux内核的标准API。

看见https://www.kernel.org/doc/htmldocs/kernel-api/ch02s03.html

set_bit  Atomically set a bit in memory
clear_bit  Clears a bit in memory
change_bit  Toggle a bit in memory
test_and_set_bit  Set a bit and return its old value
test_and_clear_bit  Clear a bit and return its old value
test_and_change_bit  Change a bit and return its old value
test_bit  Determine whether a bit is set

注意:这里整个操作在一个步骤中完成。因此,即使在SMP计算机上,这些都保证是原子的,并且是有用的以保持处理器之间的一致性。