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

Question

In below lines:

//Folder.Attributes = FileAttributes.Directory | FileAttributes.Hidden | FileAttributes.System | FileAttributes.ReadOnly;
Folder.Attributes |= FileAttributes.Directory | FileAttributes.Hidden | FileAttributes.System | FileAttributes.ReadOnly;


Folder.Attributes |= ~FileAttributes.System;
Folder.Attributes &= ~FileAttributes.System;

What does |= (single pipe equal) and &= (single ampersand equal) mean in C#?

I want to remove system attribute with keeping the others...

Solution 1

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

Solution 2

• | 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

Solution 3

I want to remove system attribute with keeping the others..

You can do this like so:

Folder.Attributes ^= FileAttributes.System;

Comments

[IronMensan]

x = x | (y); is a better way to describe it because x |= y + z; is not the same as x = x | y + z;

[SilverLight]

thanks for answers / but for my purpose(removing system attribute) which one should i use (|= or &=)?

[George Duckett]

@LostLord: Folder.Attributes &= ~FileAttributes.System;

[GameZelda]

I think you want to use a XOR instead of an AND for this.

[SilverLight]

a bit confused / ~is necessary or not

[MordechayS]

@LostLord The two methods are analogous as far as I'm aware

[silkfire]

What does it mean that a is only evaluated once? Why would it be evaluated more times than that?

[Polluks]

@silkfire This is called short-circuit evaluation, see en.wikipedia.org/wiki/Short-circuit_evaluation

[silkfire]

@Polluks So basically a |= b actually means a = a || b?

[Polluks]

@silkfire Yep but don't interchange one pipe and two pipes.

[Chronicle]

@ChrisS ^= bit will set the bit if it was not already set, &= ~bit does not set it.

[John Lord]

you definitely don't want to use an xor. That would put it back if it was gone.

[O. R. Mapper]

@Polluks: I fail to understand your comment about one and two pipes - I think using two pipes instead of one was silkfire's entire point in that last comment. Also, I am not convinced the statement "except that a is evaluated only once` does indeed refer to short-circuit evaluation, as short-circuit evaluation does not change the evaluation of the first operand ( a ), but might skip the evaluation of the second operand ( b ).

[Evgeni Nabokov]

@silkfire it means a "not lazy" ||, i.e. both parts of the expression are always calculated, whereas 2nd part of a standard "lazy" || may not be calculated.