英文:
How to swap specific bits between two bytes based on bit map?
问题
<pre>例如：如果 A -> 00001010(10)，B -> 01000001(65)，C -> 00011010(26)(位图)因此在交换特定位后，新的 A(A') -> 00000000，新的 B(B') -> 01001011解释：A -> 00001010C -> 00011010(交换 A 和 B 的特定位，如果 C 的特定位为1，即第4位、第5位、第7位（从左到右的顺序）B -> 01000001----------------A' -> 00000000B' -> 01001011</pre>
英文:
<pre>
For example:
If A -> 00001010(10), B-> 01000001(65), C -> 00011010(26)(Bit Map)
So after swapping specific bits, New A(A') -> 00000000 and
New B(B') -> 01001011
Explanation:A -> 00001010C -> 00011010(Swap bits b/w A and B if C's specific bit is 1,I.e bit 4, 5, 7(direction from left to right)B -> 01000001----------------A' -> 00000000B' -> 01001011
</pre>
答案1
得分: 1
这很简单。将A从B中仅设置MASK位：
- 清除
A中所有的MASK位：D = A & ~MASK - 从
B中获取仅MASK位：E = B & MASK - 将所需位设置到
A：RES = D | E
在大多数编程语言中，可能如下所示：
final int a = 0b00001010;final int b = 0b01000001;final int c = 0b00011010;int aa = (a & ~c) | (b & c);int bb = (b & ~c) | (a & c);System.out.format("a: %08d\n", new BigInteger(Integer.toBinaryString(aa)));System.out.format("a: %08d\n", new BigInteger(Integer.toBinaryString(bb)));
附言： 这只是基本的位操作。
英文:
This is pretty simple. To set A from B only bits MASK:
- Clear all bits
MASKfromA:D = A & ~MASK - Get only bits
MASKfromB:E = B & MASK - Set required bits to
A:RES = D | E
In most of languages it could look like this:
final int a = 0b00001010;final int b = 0b01000001;final int c = 0b00011010;int aa = (a & ~c) | (b & c);int bb = (b & ~c) | (a & c);System.out.format("a: %08d\n", new BigInteger(Integer.toBinaryString(aa)));System.out.format("a: %08d\n", new BigInteger(Integer.toBinaryString(bb)));
P.S. This is only basic bit operations.
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