在 2 个整数之间交换不同的位 - C

Question

我需要帮助。

我需要解决在 2 个不同整数之间交换 2 个不同位的问题。

示例（将 (101) 的第3位与 (100) 的第2位交换）

将导致(001) & (110)

我的审判

void swap(unsigned int numberA, unsigned int numberB, int bitPositionA, int bitPositionB)
{
    unsigned int aShift = 1 << bitPositionA, bShift = 1 << bitPositionB;
    unsigned int bitA = numberA & aShift;
    unsigned int bitB = numberB & bShift;


    numberB &= ~bShift; // Set the bit to `0`
    numberB |= bitA;    // Set to the actual bit value

    numberA &= ~aShift; // Set the bit to `0`
    numberA |= bitB;    // Set to the actual bit value

    printf("Number[1] => %d Number => %d",numberA,numberB);
}

swap(5,4,3,2) -> Number[1] => 5 Number => 0错误输出

Answer 1

1. 你忘记了像数组一样的位是从零开始编号的，而不是从一开始。

   将您对swap的调用swap为：

    swap(5, 4, 2, 1);

2. 在新位中进行 OR 运算的代码不会将它们移动到它们应该在新数字中进入的位位置。 它们保留在它们在源编号中被拉出的位位置。

    numberB &= ~bShift; // Set the bit to `0` if(bitA) bitA = 1 << bitPositionB; numberB |= bitA; // Set to the actual bit value numberA &= ~aShift; // Set the bit to `0` if(bitB) bitB = 1 << bitPositionA; numberA |= bitB; // Set to the actual bit value

Answer 2

void swap(unsigned int numberA, unsigned int numberB, int bitPositionA, int bitPositionB)
{
    unsigned int aShift = 1 << bitPositionA-1, bShift = 1 << bitPositionB-1;
    unsigned int bitA = numberA & aShift;
    unsigned int bitB = numberB & bShift;


    numberB &= ~bShift; // Set the bit to `0`
    numberB |= bitA >> (bitPositionB-1);    // Set to the actual bit value

    numberA &= ~aShift; // Set the bit to `0`
    numberA |= bitB >> (bitPositionA-1);    // Set to the actual bit value

    printf("Number[1] => %02X Number => %02X \n",numberA,numberB);
}