问题描述
我必须实现一个方法写一个字节到 ostream 对象。让我们调用这个 ostream object strobj 。我也有一个位缓冲区用于存储一个字节的数据，我们称之为：
I have to implement a method that writes a byte to an ostream object. Let's just called this ostream object strobj. I also have a bit buffer used to store a byte of data, let's call it:
char &bitter;
在我的头文件中，我有：
In my header file, I have this:
void writeThisByte(int sumInt);
指令说，我必须写入int的较低有效字节 ostream 对象，即 strobj ;
The instructions says that I have to write the "less significant byte" of the int being passed in to an ostream object, i.e. strobj;
我很困惑最低有效字节的含义。
However, I am confused on exactly what the least significant byte means.
这是否意味着你正在检查 sumInt == 1 ？如果是这样，你能像这样写入 ostream 吗？
Does this mean you are checking to see whether sumInt == 1? And if it is, do you write it to the ostream like this?
strobj.write(&bitter, 1);
我不确定：（
推荐答案
想象一个32位整数，其中每个位可以是0或1，例如：
Imagine a 32-bit integer, where each bit can be 0 or 1, say:
011101010101010101010101010110110
^^^^^^^^
最低有效字节
这可能更容易理解，如果你采取一个十进制的例子：给定数字291023，'3'是最小的 -
It's probably easier to understand if you take a decimal example: given the number 291023, the '3' is the least-significant digit, because if you change it you have the least effect on the overall number.
为了得到最低有效字节，只需按位AND就可以将较大的 int 使用0xFF十六进制或255十进制（1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255） - 将清除所有更重要的位， 8个最低有效位...
To get the least significant byte, just bitwise-AND the larger int with 0xFF hex or 255 decimal (1+2+4+8+16+32+64+128=255)- that will clear out all the more-signficant bits and preserve the 8 least significant bits...
int x = ...whatever...;
unsigned char least_significant_bits = x & 255;
效果如下：
011101010101010101010101010110110 // x
000000000000000000000000011111111 //255
result:
000000000000000000000000010110110 // x & 255
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