结构到位 C++

Question

我正在学习 C++，我想知道是否可以将一个结构对象分解成一个位序列？

// The task is this! I have a structure
struct test {
  // It contains an array
    private:
    int arr [8];

    public:
    void init () {
      for (int i = 0; i <8; i ++) {
        arr [i] = 5;
     }

    }
};


int main () {
  // at some point this array is initialized
    test h;
    h.init ();

    // without referring to the arr field and its elements, we must convert the structure to this format
    // we know that int is stored there, and these are 32 bits -> 00000000 00000000 00000000 00000101. 00000000 00000000 00000000 00000101. - and there are 8 such pieces by number
    // elements in the array

  return -1;
}

嗯，我们也知道数组的大小。我们需要将结构对象转换为位序列：

00000000000000000000000000000101000000000000000000000000000001010000000000000000000000000000010100000000000000000000000000000101000000000000000000000000000001010000000000000000000000000000010100000000000000000000000000000000010100000000000000000000000000000101

Answer 1

将数字转换为位串的标准答案是给你的std::bitset。我将使用更底层的方法。并使用位掩码和 & 操作来屏蔽掉单个位，然后将相应的字符分配给结果字符串。

使用位和运算符以及局部 AND 运算屏蔽工作

Bit Mask      AND
0   0         0
0   1         0
1   0         0
1   1         1

你看，0和1是0。1和1是1。

这允许我们访问一个字节中的一个位。

Byte: 10101010
Mask: 00001111
--------------
      00001010

我们将使用这个机制。

但是，我无法想象这是家庭作业，因为需要通过从外部访问结构来进行肮脏的 reintepret_cast。

无论如何。让我向您展示这个解决方案。

我觉得非常丑。

#include <iostream>
#include <bitset>

// The task is this! I have a structure
struct test {
    // It contains an array
private:
    int arr[8];

public:
    void init() {
        for (int i = 0; i < 8; i++) {
            arr[i] = 5;
        }
    }
};

// Convert an int to a string with the bit representaion of the int
std::string intToBits(int value) {

    // Here we will store the result
    std::string result{};

    // We want to mask the bit from MSB to LSB
    unsigned int mask = 1<<31;

    // Now we will work on 4 bytes with 8bits each
    for (unsigned int byteNumber = 0; byteNumber < 4; ++byteNumber) {
        for (unsigned int bitNumber = 0; bitNumber < 8; ++bitNumber) {

            // Mask out bit and store the resulting 1 or 0 in the string
            result += (value & mask) ? '1' : '0';

            // Next mask
            mask >>= 1;
        }
        // Add a space between bytes
        result += ' ';
    }
    // At the end, we do want to have a point
    result.back() = '.';
    return result;
}

int main() {
    // At some point this array is initialized
    test h;
    h.init();

    // Now do dirty, ugly, and none compliant type cast
    int* test = reinterpret_cast<int*>(&h);
 
    // Convert all bytes and show result
    for (unsigned int k = 0; k < 8; ++k) 
        std::cout << intToBits(test[k]) << ' ';

    return 0;
}

结构到位 C++

Struct to bits c++

c++

struct

bit

bitset