31 在这个方法中代表什么？

Question

I have this function to find log of base 2 but I dont understand the meaning of the 31.

  public static void loggg(int n){
        int result = 0;
        for(int i = 0; (n << i & 1 << 31) == 0; i++){
            result = (31-i-1);
            System.out.println( "result " +result);
        }
}

this is my output. it seems like its a max number.

result 30
result 29
result 28
result 27
result 26
result 25
result 24
result 23
result 22
result 21
result 20
result 19
result 18
result 17
result 16
result 15
result 14
result 13
result 12
result 11
result 10
result 9
result 8
result 7
result 6
result 5
result 4

Answer 1

(n << i & 1 << 31) == 0

is an obscure way of writing

(n << i) >= 0

1 << 31 is the binary number 10000000 00000000 00000000 00000000 . It's 1, shifted left 31 times.

X & 1 << 31 is a mask, checking if bit 31 is set to 1 in X .

Bit 31 in a signed 32-bit integer is the sign bit. So X & 1 << 31 == 0 is just checking if X is non-negative.

Answer 2

You said this finds the log to the base 2. But since you are working with integers it really only finds the log of the highest power of 2 <= n

An easier way (but not as interesting) is to simply do the following.

System.out.println(31-Integer.numberOfLeadingZeros(n));