使用1按位设置字节的所有位 [英] Using 1 to bitwise set all bits of a byte

问题描述

我目前像这样将值硬编码为255

I currently hardcode a value to 255 like so

uint8_t outStates       = 0xFF;

但是我有一个更干净使用的宏

But I have a macro that would be cleaner to use

#define TCA9555_LOW (1)

我想到的唯一方法就是这种方式，但是阅读效率低下且难看，hich不是我想要的

And the only way I would think of to do it would be this way, but is seems inefficient and uglier to read, hich is not what I want

for(int i = 0; i < 8; i++)
{
    outStates |= (TCA9555_LOW << i);
}

是否有一种更漂亮的方法，可以使用易于读取的值为1的宏一次设置所有位?

Is there a prettier way to set all bits at once, using an easy to read macro of value 1 ?

非常感谢！

推荐答案

您希望将每个位设置为 TCA9555_LOW 的值而不会循环.为此，您可以使用以下代码:

You want to to set each bit to the value of TCA9555_LOW without looping. To achieve that, you can use the following:

uint8_t outStates =
     ( TCA9555_LOW << 7 )
   | ( TCA9555_LOW << 6 )
   | ( TCA9555_LOW << 5 )
   | ( TCA9555_LOW << 4 )
   | ( TCA9555_LOW << 3 )
   | ( TCA9555_LOW << 2 )
   | ( TCA9555_LOW << 1 )
   | ( TCA9555_LOW << 0 );

---

但是，如果我们考虑一下，只有两种可能的结果.以下是等效的:

---

But if we think about it, there's only two possible outcomes. The following would be equivalent:

uint8_t outStates = TCA9555_LOW ? 0xFF : 0x00;

---

最后，在相同的基础上，我们可以使用以下内容:

---

Finally, on the same basis, we could use the following:

uint8_t outStates = 0;
if (TCA9555_LOW)
   outStates = ~outStates;

---

在前两种情况下，编译器应将表达式折叠为单个常量.甚至第三.

---

In the first two cases, the compiler should fold the expression into a single constant. Maybe even the third.

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