问题描述
限时送ChatGPT账号..谁能告诉我这里做了什么:
Can someone tell me what is being done here:
Const uint32_t goodguys = 0x1 << 0
我假设它是 C++ 并且它正在为一个组分配一个标签,但我从未见过这样做过.我是一个自学的客观 c 家伙,这对我来说看起来很陌生.
I'm assuming it is c++ and it is assigning a tag to a group but I have never seen this done. I am a self taught objective c guy and this just looks very foreign to me.
推荐答案
好吧,如果在您发布的行之后还有更多类似这样的行,那么它们可能是 位掩码
.
Well, if there are more lines that look like this that follow the one that you posted, then they could be bitmasks
.
例如,如果您有以下内容:
For example, if you have the following:
const uint32_t bit_0 = 0x1 << 0;
const uint32_t bit_1 = 0x1 << 1;
const uint32_t bit_2 = 0x1 << 2;
...
然后您可以将按位 &
运算符与 bit_0
、bit_1
、bit_2
、... 和另一个数字,以便查看另一个数字中的哪些位被打开.
then you could use use the bitwise &
operator with bit_0
, bit_1
, bit_2
, ... and another number in order to see which bits in that other number are turned on.
const uint32_t num = 5;
...
bool bit_0_on = (num & bit_0) != 0;
bool bit_1_on = (num & bit_1) != 0;
bool bit_2_on = (num & bit_2) != 0;
...
所以你的 0x1
只是一种指定 goodguys
是位掩码的方法,因为十六进制的 0x
指示符表明代码正在专门考虑位,而不是十进制数字.然后 <<0
用于准确更改位掩码的掩码内容(您只需将 0
更改为 1
、2
等即可.).
So your 0x1
is simply a way to designate that goodguys
is a bitmask, because the hexadecimal 0x
designator shows that the author of the code is thinking specifically about bits, instead of decimal digits. And then the << 0
is used to change exactly what the bitmask is masking (you just change the 0
to a 1
, 2
, etc.).
这篇关于这行代码有什么作用?const uint32_t goodguys = 0x1 <<0的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
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