it's bitshift operation.
a byte for example is
01101001
if this byte is ouput[5] then
01101001 >> 3 = 00001101 (shiftet 3 bit on the left)
then biwise and:
00001101 & 2 = 00001101 & 00000010 (this is 2 in binary) = 0
maybe i'm wrong, but this 2 operation ban be just one:
01101001 & 00010000 (staring byte and 2^4=16)
so
ouput[5] & 16 == ((outbuf[5] >> 3) & 2)
as you can see, all this complexity is just to know if the 5° bit from the right is 1 or 0 