Can someone explain the following line of code? Particularly, I don't get what (short) x & 0x3FF does?
int num = ... //some number.
return (short) num & 0x3FF;5 Answers
It zeros out the top bits of the number, such that the result is always between 0 and 1023. It's essentially the same thing as modulo(num, 1024) (for positive values of num).
Without seeing a broader context it's impossible to know why this is here, but that's what it does.
00x3FF is the number 1111111111 in binary, which means that the bitwize AND with it would give you the last 10 bits of num.
Converting hex to binary, 0x3FF == 0b1111111111.
The & performs bitwise AND, so it will only keep the low bits if they were set to be on.
This guarantees that the answer is no larger than 0x3FF == 1023, so the answer is saved into a short, since we know it'll fit in one.
The java operator & is a "bitwise AND", meaning each bit of the two operands is ANDed together, leaving a 1 if both corresponding bits are 1.
0x3ff is binary 1111111111, so ANDing with that will mask all but the lowest ten bits.
The cast to short (a 16-bit number format) has no effect.
It retains only the right-most 10 bits.
