如何阅读一个unsigned int的特定位 [英] How to read specific bits of an unsigned int

问题描述

我有一个uint8_t有，我需要读/写特定位。我怎么会去这样做。具体是什么我的意思是，我需要写，再后来读了前7位的一个值和另一个值的最后一位。

I have an uint8_t and I need to read/write to specific bits. How would I go about doing this. Specifically what I mean is that I need to write and then later read the first 7 bits for one value and the last bit for another value.

编辑：忘了指定，我将这些设置为大端

edit: forgot to specify, I will be setting these as big endian

推荐答案

您正在寻找的 bitmasking 的。学习如何用C的位运算符：〜， | ，＆安培; ， ^ 等将是巨大的帮助，我建议你看看他们。

You're looking for bitmasking. Learning how to use C's bitwise operators: ~, |, &, ^ and so on will be of huge help, and I recommend you look them up.

否则 - 要读出的最低显著位？

Otherwise -- want to read off the least significant bit?

uint8_t i = 0x03;

uint8_t j = i & 1; // j is now 1, since i is odd (LSB set)

和设置呢？

uint8_t i = 0x02;
uint8_t j = 0x01;

i |= (j & 1); // get LSB only of j; i is now 0x03

希望我的七个最显著位设置到j的七个最显著位？

Want to set the seven most significant bits of i to the seven most significant bits of j?

uint8_t j = 24; // or whatever value
uint8_t i = j & ~(1); // in other words, the inverse of 1, or all bits but 1 set

想读出我的这些位？

Want to read off these bits of i?

i & ~(1);

要（从零，其中0表示LSB索引）我的位？

Want to read the Nth (indexing from zero, where 0 is the LSB) bit of i?

i & (1 << N);

和设置呢？

i |= (1 << N); // or-equals; no effect if bit is already set

这些技巧会在pretty方便你学习C

These tricks will come in pretty handy as you learn C.

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