英文:
Get / Set only specific bits in a long
问题
我有一个包含多个字段的48位结构，我希望能够读取和操作这些字段。我认为我已经理解了获取部分，但在设置部分遇到了困难。以下是一些示例代码：
问题是，在设置器中，如何只设置特定的位而不影响其他位。我理解这可以通过位移和一些掩码操作来实现。
例如：
public long data = 0x00;
public byte val1
{
get => (byte)(data >> 0 & 3); //位 0-1
}
public byte val2 //位 2-9
{
get => (byte)(data >> 2 & 0x3FC); //我认为这个可以工作？
set
{
// ?
}
}
基本上，我正在尝试从这个长整型中读取/插入一个值（int16、int32、byte等）。我尝试使用 BitVector，但最大的部分是16位，而其中一些字段的大小超过了这个（24位）。
英文:
I have a 48 bit struct with several fields that I want to be able to read and manipulate. I think i got the getter part, but I am struggling with the setter. Here is some sample code:
Question is, for the setter, how can I just set specific bits without affecting the others. My understanding is that accomplished by bit shifting and some mask operations.
For example:
public long data = 0x00;
public byte val1
{
get => (byte)(data >>0 & 3); //Bit 0-1
}
public byte val2 //Bit 2-9
{
get => (byte)(data >> 2 & 0x3FC); //I think this works?
set
{
// ?
}
}
Basically I am trying to read / insert a value (int16, int32, byte, etc) into that long. I tried using BitVector but the biggest section is 16 bits and a couple of the fields are bigger than that. (24 bits).
答案1
得分: 1
Sure, here's the translated code:
未经测试但：
```cpp
const long mask = 0x3FC;
long i64 = value;
data = (data & ~mask) | (i64 << 2);
<details>
<summary>英文:</summary>
untested but:
const long mask = 0x3FC;
long i64 = value;
data = (data & ~mask) | (i64 << 2);
</details>
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