c++11 - 如何使用按位运算将 3 个整数值编码为 uint16_t？

Question

我想将 3 个无符号整数值存储到 uint16_t 中变量通过按位运算并使用按位运算读回。以下是我的程序:
代码:

#include <iostream>

uint16_t Write(unsigned int iVal1, unsigned int iVal2, unsigned int iVal3) {
    // iVal1 should go into the first 8 bits [iVal1 value ranges from 0 to 173]
    // iVal2 should go into the 6 bits after that [iVal2 value ranges from 0 to 63]
    // iVal3 should go into the 2 bits after that [iVal3 value ranges from 0 to 3]
    // Is the below way of writing the bits correct?
    return (static_cast<uint16_t>(iVal1)<<8) + (static_cast<uint16_t>(iVal2)<<6) + (static_cast<uint16_t>(iVal3)<<2);
}

unsigned int ReadVal1(const uint16_t theNumber) {
    // ival1 is the first 8 bits
    uint16_t check1 = 255;
    return (theNumber>>8)&check1;
}

unsigned int ReadVal2(const uint16_t theNumber) {
    // ival2 is the 6 bits after that
    uint16_t check2 = 63;
    return (theNumber>>3)&check2;
}

unsigned int ReadVal3(const uint16_t theNumber) {
    // ival3 is the last 2 bits
    uint16_t check3 = 3;
    return (theNumber>>1)&check3;
}

int main() {
    std::cout << "Main started" << std::endl;

    unsigned int iVal1 = 191;
    unsigned int iVal2 = 28;
    unsigned int iVal3 = 3;

    const uint16_t theNumber = Write(iVal1, iVal2, iVal3);

    std::cout << "The first 8 bits contain the number: " << ReadVal1(theNumber) << std::endl;
    std::cout << "Then after 6 bits contain the number: " << ReadVal2(theNumber) << std::endl;
    std::cout << "Then after 2 bits contain the number: " << ReadVal3(theNumber) << std::endl;
}

在上面的程序中，以下是需要编码的 3 个无符号整数的范围。

`iVal1` ranges from `0 to 173`. So its well within 8 bits.
`iVal2` ranges from `0 to 63`. So its well within 6 bits.
`iVal3` ranges from `0 to 3`. So its well within 2 bits.

问题:
我认为我在函数中写入值的方式 Write是错的。什么是正确的方法？
首先，我正在寻找一个很好的解释，说明使用按位运算的编码是如何工作的，尤其是在我上面程序的目标上下文中。
我相信我读取函数中值的方式 ReadVal1 , ReadVal2和 ReadVal3是正确的。我已经想出了如何读回似乎很容易的值的技巧。但是，我无法很好地理解如何使用按位运算正确编码值的逻辑。
C++ 编译器:
我正在使用 C++11 编译器

Answer 1

移位整数的位数不应该取决于被移位的整数的大小，而是取决于它之后(向右)的所有整数的大小。这里有一些 ASCII 艺术来说明这个原理:

                                +---+---+---+---+---+---+---+---+
                                ‖i1 |i1 |i1 |i1 |i1 |i1 |i1 |i1 ‖ 8 bit
                                ‖ 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 ‖
                                +---+---+---+---+---+---+---+---+
                                 _______________________________/
                                /                     
                                |       +---+---+---+---+---+---+
                                |       ‖i2 |i2 |i2 |i2 |i2 |i2 ‖ 6 bit
                                |       ‖ 5 | 4 | 3 | 2 | 1 | 0 ‖
                                |       +---+---+---+---+---+---+
                                |                       ________/
                                |                      /
                                |                      |+---+---+
                                |                      |‖i3 |i3 ‖ 2 bit
                                |                      |‖ 1 | 0 ‖
                                |                      |+---+---+
                                |                      \        |
                                |<<(6+2)                |<<2    |<<0
                                v                       v       v
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
‖ F | E | D | C | B | A | 9 | 8 ‖ 7 | 6 | 5 | 4 | 3 | 2 ‖ 1 | 0 ‖
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+
‖i1 |i1 |i1 |i1 |i1 |i1 |i1 |i1 ‖i2 |i2 |i2 |i2 |i2 |i2 ‖i3 |i3 ‖
‖ 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 ‖ 5 | 4 | 3 | 2 | 1 | 0 ‖ 1 | 0 ‖
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+