Bit Number to Byte Value conversion (AVR Docs) I have hard time to understand what this doc tells me: https://www.microchip.com/webdoc/AVRLibcReferenceManual/FAQ_1faq_use_bv.html

I has a macro called _BV that 


  convert a bit number (usually within a byte register) into a byte value


And the example tells this


  _BV(3) => 1 &lt;&lt; 3 => 0x08


I am kinda used to bits and bytes etc. But I don't understand what the 3 in this example is? It says it converts a bit number but 3 is not binary it is an integer. And the macro resolves it to 0x08 which is the value of 8 in decimal.

I am very confused. I hope someone can explain to me what is happening here!
The bits within a byte are numbered 0 through 7 from right to left. The
rightmost (least significant) is bit 0, next is bit 1... and the
leftmost (most significant) is bit 7.

The expression _BV(3) builds a byte where bit 3 is set (is 1) and all
other bits are clear (are 0). In binary it looks like this:

position:   7  6  5  4  3  2  1  0
bit value:  0  0  0  0  1  0  0  0


The macro expands to 1&lt;&lt;3, which takes the integer 1 (0b00000001 in
binary) and shifts all its bits by three positions to the left. The
result can be written 0b00001000 in binary, 0x08 in hexadecimal, and 8
in decimal.