bitwise shifts
Task condition:
Implement the
flipBitmethod, which changes the value of one bit of a given integer to the opposite.Let's agree that the bits are numbered from the lowest (index 1) to the highest (index 32).
And here is his solution:
public static int flipBit(int value, int bitIndex) {
return value ^ (1 << bitIndex-1);
}
Question: I can't understand what is happening here:
(1 << bitIndex-1)
Explain pozh-a.
3 answers
Question: I can't understand what is happening here: (1
There is a shift of bits to the left (or stupidly raising the two to the power) bitIndex-1
In fact, you can see more details here
In short, 1 can be represented roughly as 00000001
<< - bitwise left shift
As a result, for example, if bitIndex is 3, then 1 << 3 (we shift one by three digits to the left) turns out
00001000 // 8
And when bitIndex - 1 it turns out accordingly:
bitIndex - 1 → 1 << 2 → 00000100 // 4
Although in the end, what is the value in the decimal system is not important, but it is simple, for the record
Question: I can't understand what is happening here:
(1 << bitIndex-1)
Here, the bits in 1 are shifted to the left by bitIndex-1 positions.
The bits are shifted to the left, and the new bits are replaced by zeros. In fact, this is the raising of a deuce to a power.
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