問(wèn)題描述
我必須翻轉(zhuǎn)整數(shù)二進(jìn)制表示中的所有位.給定:
I have to flip all bits in a binary representation of an integer. Given:
10101
輸出應(yīng)該是
01010
當(dāng)與整數(shù)一起使用時(shí)，完成此操作的位運(yùn)算符是什么?例如，如果我正在編寫(xiě)像 int flipBits(int n); 這樣的方法，那么主體中會(huì)發(fā)生什么?我只需要翻轉(zhuǎn)數(shù)字中已經(jīng)存在的內(nèi)容，而不是整數(shù)中的所有 32 位.
What is the bitwise operator to accomplish this when used with an integer? For example, if I were writing a method like int flipBits(int n);, what would go in the body? I need to flip only what's already present in the number, not all 32 bits in the integer.
推薦答案
~ 一元運(yùn)算符是按位取反.如果您需要的位數(shù)少于 int 中的位數(shù)，那么您需要在事后使用 & 對(duì)其進(jìn)行屏蔽.
The ~ unary operator is bitwise negation. If you need fewer bits than what fits in an int then you'll need to mask it with & after the fact.
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