What Does |= (Single Pipe Equal) and &=(Single Ampersand Equal) Mean

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

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

What does the |= Operator mean?

You know how x += 1 means x = x + 1, well x |= 1 means x = x | 1. Of course | means bitwise OR.

What does |= operator do?

dirtyFlag |= 0x0001 is equivalent to dirtyFlag = dirtyFlag | 0x0001. The | operator is the bitwise OR operator. In your case, it sets the lowest binary bit. Some further examples:

1 | 2 = 3 (0001 | 0010 = 0011)
2 | 4 = 6 (0010 | 0100 = 0110)
5 | 1 = 5 (0101 | 0001 = 0101)

What does & (single ampersand), and | (single pipe) operator do in comparison operation?

Those are both Bitwise Operators, which let you manipulate the actual data bits.

The information below was copied as-is from the official documentation:

Bitwise AND Operator

The bitwise AND operator (&) combines the bits of two numbers. It returns a new number whose bits are set to 1 only if the bits were equal to 1 in both input numbers:

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Bitwise OR Operator

The bitwise OR operator (|) compares the bits of two numbers. The operator returns a new number whose bits are set to 1 if the bits are equal to 1 in either input number:

What is the difference between the | and || or operators?

Just like the & and && operator, the double Operator is a "short-circuit" operator.

For example:

if(condition1 || condition2 || condition3)

If condition1 is true, condition 2 and 3 will NOT be checked.

if(condition1 | condition2 | condition3)

This will check conditions 2 and 3, even if 1 is already true. As your conditions can be quite expensive functions, you can get a good performance boost by using them.

There is one big caveat, NullReferences or similar problems. For example:

if(class != null && class.someVar < 20)

If class is null, the if-statement will stop after class != null is false. If you only use &, it will try to check class.someVar and you get a nice NullReferenceException. With the Or-Operator that may not be that much of a trap as it's unlikely that you trigger something bad, but it's something to keep in mind.

No one ever uses the single & or | operators though, unless you have a design where each condition is a function that HAS to be executed. Sounds like a design smell, but sometimes (rarely) it's a clean way to do stuff. The & operator does "run these 3 functions, and if one of them returns false, execute the else block", while the | does "only run the else block if none return false" - can be useful, but as said, often it's a design smell.

There is a Second use of the | and & operator though: Bitwise Operations.

What does |= mean? (pipe equal operator)

|= reads the same way as +=.

notification.defaults |= Notification.DEFAULT_SOUND;

is the same as

notification.defaults = notification.defaults | Notification.DEFAULT_SOUND;

where | is the bit-wise OR operator.

All operators are referenced here.

A bit-wise operator is used because, as is frequent, those constants enable an int to carry flags.

If you look at those constants, you'll see that they're in powers of two :

public static final int DEFAULT_SOUND = 1;
public static final int DEFAULT_VIBRATE = 2; // is the same than 1<<1 or 10 in binary
public static final int DEFAULT_LIGHTS = 4; // is the same than 1<<2 or 100 in binary

So you can use bit-wise OR to add flags

int myFlags = DEFAULT_SOUND | DEFAULT_VIBRATE; // same as 001 | 010, producing 011

so

myFlags |= DEFAULT_LIGHTS;

simply means we add a flag.

And symmetrically, we test a flag is set using & :

boolean hasVibrate = (DEFAULT_VIBRATE & myFlags) != 0;

Understanding the behavior of a single ampersand operator (&) on integers

Compare the binary representations of each of those.

    110 &     010 =     010
   1010 &    0101 =    0000
  10100 &   11001 =   10000
1111011 & 0010100 = 0010000

In each case, a digit is 1 in the result only when it is 1 on both the left AND right side of the input.

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