什么是到0xFF吗？ [英] What does AND 0xFF do?

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<击> 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 tristopiaPatrick 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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