问题描述
在以下code：
short = ((byte2 << 8) | (byte1 & 0xFF))
什么是＆放大器的目的;为0xFF ？
因为其他somestimes我把它写成：
What is the purpose of &0xFF?Because other somestimes I see it written as:
short = ((byte2 << 8) | byte1)
这似乎很好地工作呢？
And that seems to work fine too?
推荐答案
与安定 0xFF的只留下至少显著字节整数。例如，为了获得在短S中的第一个字节，你可以写 S＆安培; 0xFF的。这通常被称为掩蔽。如果字节1 是一个单字节类型（如 uint8_t有）或已小于256（，因此是除了至少显著字节全零）没有必要掩盖了更高位，如他们已经为零。
Anding an integer with 0xFF leaves only the least significant byte. For example, to get the first byte in a short s, you can write s & 0xFF. This is typically referred to as "masking". If byte1 is either a single byte type (like uint8_t) or is already less than 256 (and as a result is all zeroes except for the least significant byte) there is no need to mask out the higher bits, as they are already zero.
请参阅 tristopia 帕特里克SCHLÜTER的答案时，你可能会签订各类合作下方。在做位操作，我建议只与无符号类型的工作。
See Patrick Schlüter's answer below when you may be working with signed types. When doing bitwise operations, I recommend working only with unsigned types.
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