如何使用按位运算将 3 个整数值编码为 uint16_t？

Question

我想通过按位运算将 3 个无符号整数值存储到 uint16_t 变量中，然后使用按位运算将它们读回。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;
}

上面程序中需要编码的3个无符号整数的取值范围如下

`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 Write 中编写值的方式是错误的。正确的做法是什么？

主要是，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，尤其是在我上面的程序目标的上下文中。

我相信我读取函数 ReadVal1、ReadVal2 和 ReadVal3 中的值的方式是正确的。我已经想出了如何回读看似简单的值的技巧。但是，我不太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

整数移位的位数不应该取决于被移位的整数的大小，而是取决于在它之后的所有整数的大小（到正确的）。这里有一些 ASCII 艺术来说明原理：

                                +---+---+---+---+---+---+---+---+
                                ‖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 ‖
+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+---+

如何使用按位运算将 3 个整数值编码为 uint16_t？

How to encode 3 integer values into an uint16_t using bitwise operations?

bit-manipulation

bit-shift

bitwise-operators

c++11