What does |= (single pipe equal) and &=(single ampersand equal) mean

• | is bitwise or
• & is bitwise and

a |= b is equivalent to a = a | b except that a is evaluated only once
a &= b is equivalent to a = a & b except that a is evaluated only once

In order to remove the System bit without changing other bits, use

Folder.Attributes &= ~FileAttributes.System;

~ is bitwise negation. You will thus set all bits to 1 except the System bit. and-ing it with the mask will set System to 0 and leave all other bits intact because 0 & x = 0 and 1 & x = x for any x

They're compound assignment operators, translating (very loosely)

x |= y;

into

x = x | y;

and the same for &. There's a bit more detail in a few cases regarding an implicit cast, and the target variable is only evaluated once, but that's basically the gist of it.

In terms of the non-compound operators, & is a bitwise "AND" and | is a bitwise "OR".

EDIT: In this case you want Folder.Attributes &= ~FileAttributes.System. To understand why:

• ~FileAttributes.System means "all attributes except System" (~ is a bitwise-NOT)
• & means "the result is all the attributes which occur on both sides of the operand"

So it's basically acting as a mask - only retain those attributes which appear in ("everything except System"). In general:

• |= will only ever add bits to the target
• &= will only ever remove bits from the target