Question

If we have a decimal value: 123
and its binary version: 01111011

How can I get four leftmost and the four rightmost bits from this byte into 2 separate int variables?

I mean:

int a = 7;  // 0111 (the first four bits from the left)
int b = 11; // 1011 (the first four bits from the right)

Much appreciated!

Answer 1

int x = 123;
int low = x & 0x0F;
int high = (x & 0xF0) >> 4;

This is called masking and shifting. By ANDing with 0xF (which is binary 00001111) we remove the higher four bits. ANDing with 0xF0 (which is binary 11110000) removes the lower four bits. Then (in the latter case), we shift to the right by 4 bits, in effect, pushing away the lower 4 bits and leaving only what were the upper 4 bits.

As @owlstead says in the comments below, there’s another way to get the higher bits. Instead of masking the lower bits then shifting, we can just shift.

int high = x >> 4;

Note that we don’t need to mask the lower bits since whatever they were, they’re gone (we’ve pushed them out). The above example is clearer since we explicitly zero them out first, but there’s no need to do so for this particular example.

But to deal with numbers bigger than 16 bits (int is usually 32 bits), we still need to mask, because we can have the even higher sixteen bits getting in the way!

int high = (x >> 4) & 0x0F;