Question

如果我有 by,那么这种方法将如何在某个职位上检索一个轨道?

我知道这一点,我认为这行不通。

public byte getBit(int position) {
    return (byte) (ID >> (position - 1));
}

www.un.org/Depts/DGACM/index_french.htm

Answer 1

public byte getBit(int position)
{
   return (ID >> position) & 1;
}

按职位分列的正确身份证明文件将使得在最远的地方对编号正确。将“双向”和“<条码”与“;

position = 2
ID = 5 = 0000 0101 (in binary)
ID >> position = 0000 0001

0000 0001 & 0000 0001( 1 in binary ) = 1, because the furthest right bit is set.

Answer 2

你们想作一面掩盖,做两点。这最终将非常接近你——利用轮班确定适当的比照,使用<代码>&进行双向交易。

So

 return ((byte)ID) & (0x01 << pos) ;

where pos has to range between 0 and 7. If you have the least significant bit as "bit 1" then you need your -1 but I d recommend against it -- that kind of change of position is always a source of errors for me.

Answer 3

a. 进入绝缘体

 return ((num >> (n-1)) & 1);

Answer 4

In Java the following works fine:

if (value << ~x < 0) {
   // xth bit set
} else {
   // xth bit not set
}

int或long。 (也不一定相同)。

www.un.org/Depts/DGACM/index_spanish.htm 非Java方案设计员的告诫:在 Java的前面的表述作品,因为用该语言,轨道转换操作者只适用于右侧操作体最低的5条(或6条,long)。默示将这一表述转化为<代码>、数值和带;< (~x & 31)(或、数值和带;< (~x & 63)。

<>strong>Javascript:它也在javascript(如java,只有最低5倍的转移计数适用)。在javascript中, 编号为32-bit。

特别是在C,消极的转变是援引未经界定的行为,因此这一检验必然取得一定的工作(但取决于您汇编者/加工者的具体组合)。

Answer 5

采取下列步骤:

Create bitMask
Do & (AND) bitwise operation

In the number 5, the positions are from right to left (as given by index) Number (5) = |0|1|0|1| Positions = |3|2|1|0|

现在,你想在第3位(指数=2)获得比值。

int bitMask = 1 << position
int newNumber = bitMask & number

i.e:

int number = 5
int position = 2
int bitMask = 1 << position
// bitMask value becomes here:
// 1 << 2 (0001 << 2) becomes 0100
int newNumber = bitMask & number
// do AND (& bitwise) operation with bitMask and number
0100 & 0101 becomes 0100
newNumber = 4 (0100)

你们可以恢复真实或虚假的:

return ((bitMask & number) != 0)
// means if the new number is non-zero, it means the bit as position is 1 otherwise false.

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