在 2 個整數之間交換不同的位 - C

Question

我需要幫助。

我需要解決在 2 個不同整數之間交換 2 個不同位的問題。

示例（將 (101) 的第3位與 (100) 的第2位交換）

將導致(001) & (110)

我的審判

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錯誤輸出

Answer 1

1. 你忘記了像數組一樣的位是從零開始編號的，而不是從一開始。

   將您對swap的調用swap為：

    swap(5, 4, 2, 1);

2. 在新位中進行 OR 運算的代碼不會將它們移動到它們應該在新數字中進入的位位置。 它們保留在它們在源編號中被拉出的位位置。

    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);
}