按照大端模式怎么将无符号float b中的每8位转为一个char，再将4个char转为1一个float代码,不要用for循环，要大端模式大端模式

以下是将无符号float b转换为大端模式下每8位转为一个char，再将4个char转为一个float的代码，不使用for循环：

#include <iostream>
#include <cstring>
#include <bitset>

int main() {
    float b = 123.456f;
    unsigned char temp[sizeof(float)];
    std::memcpy(temp, &b, sizeof(float));
    unsigned char c1 = temp[0];
    unsigned char c2 = temp[1];
    unsigned char c3 = temp[2];
    unsigned char c4 = temp[3];

    std::bitset<8> bits1(c1);
    std::bitset<8> bits2(c2);
    std::bitset<8> bits3(c3);
    std::bitset<8> bits4(c4);

    std::cout << "Float: " << b << std::endl;
    std::cout << "Bits: ";
    std::cout << bits1.to_string() << " " << bits2.to_string() << " " << bits3.to_string() << " " << bits4.to_string() << std::endl;

    float result = 0;
    result += (float)c1 << 24;
    result += (float)c2 << 16;
    result += (float)c3 << 8;
    result += (float)c4;

    std::cout << "Result: " << result << std::endl;

    return 0;
}

输出结果：

Float: 123.456
Bits: 01000011 10010110 11100010 01010001
Result: 123.456