存储号码部分字节，但不是从开始？(Store number in part of byte, but not from start? [closed])

存储号码部分字节，但不是从开始？(Store number in part of byte, but not from start? [closed])

我有一个字节： A ：0 B ：0 C ：0 D ：0 E ：0 F ：0 G ：0 H ：0

我想以0-31，最快的方式存储一个数字，只使用字符C，D，E，F，G的空格。换句话说，我想将第1位和第2位留空，使用位3 -7存储数字并将第8位清空。

我可以使用c |= 1 << n;设置一个字节的n位c |= 1 << n; 但我不明白如何从pos 2开始？

I have a byte: A:0 B:0 C:0 D:0 E:0 F:0 G:0 H:0

I want to store a number from 0-31, the fastest way, using only the space of characters C, D, E, F, G. In other words, I want to leave the bits 1 and 2 empty, use the bits 3-7 to store the number and have bit 8 empty.

I can set the n bit of a byte using c |= 1 << n; But I fail to understand how to make it start from pos 2?

最满意答案

假设您对可读字符的ASCII范围感兴趣（32-126），则不会有5位的有效范围来满足所需的编码。 例如：

~ (126) = 01111110 (Your Mask) = 01111100

因此， ~ （126）与|相同 （124）。

如果您已经以某种方式调整了编码并且可以确保只有低5位的数据，那么您可以简单地使用按位运算：

unsigned char a = your_data; a = (a & 0x1F) << 2; // shift the lower 5 bits over 2 bits

Assuming you are interested in the ASCII range for readable characters (32-126), you will not have the valid range in 5 bits to meet the required encoding. For example:

~ (126) = 01111110 (Your Mask) = 01111100

Thus, ~ (126) would be the same as | (124).

If you have somehow already adjusted your encoding and can be assured that you will only have data in the lower 5 bits, than you can simply use bitwise operations:

unsigned char a = your_data; a = (a & 0x1F) << 2; // shift the lower 5 bits over 2 bits