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Bit Syntax

bit_syntax.xml

## Introduction

The complete specification for the bit syntax appears in the Reference Manual.

In Erlang, a Bin is used for constructing binaries and matching binary patterns. A Bin is written with the following syntax:

`<<E1, E2, ... En>>`

A Bin is a low-level sequence of bits or bytes. The purpose of a Bin is to enable construction of binaries:

`Bin = <<E1, E2, ... En>>`

All elements must be bound. Or match a binary:

`<<E1, E2, ... En>> = Bin`

Here, Bin is bound and the elements are bound or unbound, as in any match.

A Bin does not need to consist of a whole number of bytes.

A bitstring is a sequence of zero or more bits, where the number of bits does not need to be divisible by 8. If the number of bits is divisible by 8, the bitstring is also a binary.

Each element specifies a certain segment of the bitstring. A segment is a set of contiguous bits of the binary (not necessarily on a byte boundary). The first element specifies the initial segment, the second element specifies the following segment, and so on.

The following examples illustrate how binaries are constructed, or matched, and how elements and tails are specified.

### Examples

Example 1: A binary can be constructed from a set of constants or a string literal:

`Bin11 = <<1, 17, 42>>, Bin12 = <<"abc">>`

This gives two binaries of size 3, with the following evaluations:

binary_to_list(Bin11) evaluates to [1, 17, 42]. binary_to_list(Bin12) evaluates to [97, 98, 99].

Example 2:Similarly, a binary can be constructed from a set of bound variables:

`A = 1, B = 17, C = 42, Bin2 = <<A, B, C:16>>`

This gives a binary of size 4. Here, a size expression is used for the variable C to specify a 16-bits segment of Bin2.

binary_to_list(Bin2) evaluates to [1, 17, 00, 42].

Example 3: A Bin can also be used for matching. D, E, and F are unbound variables, and Bin2 is bound, as in Example 2:

`<<D:16, E, F/binary>> = Bin2`

This gives D = 273, E = 00, and F binds to a binary of size 1: binary_to_list(F) = [42].

Example 4: The following is a more elaborate example of matching. Here, Dgram is bound to the consecutive bytes of an IP datagram of IP protocol version 4. The ambition is to extract the header and the data of the datagram:

`-define(IP_VERSION, 4). -define(IP_MIN_HDR_LEN, 5). DgramSize = byte_size(Dgram), case Dgram of <<?IP_VERSION:4, HLen:4, SrvcType:8, TotLen:16, ID:16, Flgs:3, FragOff:13, TTL:8, Proto:8, HdrChkSum:16, SrcIP:32, DestIP:32, RestDgram/binary>> when HLen>=5, 4*HLen=<DgramSize -> OptsLen = 4*(HLen - ?IP_MIN_HDR_LEN), <<Opts:OptsLen/binary,Data/binary>> = RestDgram, ... end.`

Here, the segment corresponding to the Opts variable has a type modifier, specifying that Opts is to bind to a binary. All other variables have the default type equal to unsigned integer.

An IP datagram header is of variable length. This length is measured in the number of 32-bit words and is given in the segment corresponding to HLen. The minimum value of HLen is 5. It is the segment corresponding to Opts that is variable, so if HLen is equal to 5, Opts becomes an empty binary.

The tail variables RestDgram and Data bind to binaries, as all tail variables do. Both can bind to empty binaries.

The match of Dgram fails if one of the following occurs:

The first 4-bits segment of Dgram is not equal to 4. HLen is less than 5. The size of Dgram is less than 4*HLen.

## Lexical Note

Notice that "B=<<1>>" will be interpreted as "B =< <1>>", which is a syntax error. The correct way to write the expression is: B = <<1>>.

## Segments

Each segment has the following general syntax:

Value:Size/TypeSpecifierList

The Size or the TypeSpecifier, or both, can be omitted. Thus, the following variants are allowed:

Value Value:Size Value/TypeSpecifierList

Default values are used when specifications are missing. The default values are described in Defaults.

The Value part is any expression, when used in binary construction. Used in binary matching, the Value part must be a literal or a variable. For more information about the Value part, see Constructing Binaries and Bitstrings and Matching Binaries.

The Size part of the segment multiplied by the unit in TypeSpecifierList (described later) gives the number of bits for the segment. In construction, Size is any expression that evaluates to an integer. In matching, Size must be a constant expression or a variable.

The TypeSpecifierList is a list of type specifiers separated by hyphens.

Type The most commonly used types are integer, float, and binary. See Bit Syntax Expressions in the Reference Manual for a complete description. Signedness The signedness specification can be either signed or unsigned. Notice that signedness only matters for matching. Endianness The endianness specification can be either big, little, or native. Native-endian means that the endian is resolved at load time, to be either big-endian or little-endian, depending on what is "native" for the CPU that the Erlang machine is run on. Unit The unit size is given as unit:IntegerLiteral. The allowed range is 1-256. It is multiplied by the Size specifier to give the effective size of the segment. The unit size specifies the alignment for binary segments without size.

Example:

`X:4/little-signed-integer-unit:8`

This element has a total size of 4*8 = 32 bits, and it contains a signed integer in little-endian order.

## Defaults

The default type for a segment is integer. The default type does not depend on the value, even if the value is a literal. For example, the default type in <<3.14>> is integer, not float.

The default Size depends on the type. For integer it is 8. For float it is 64. For binary it is all of the binary. In matching, this default value is only valid for the last element. All other binary elements in matching must have a size specification.

The default unit depends on the the type. For integer, float, and bitstring it is 1. For binary it is 8.

The default signedness is unsigned.

The default endianness is big.

## Constructing Binaries and Bitstrings

This section describes the rules for constructing binaries using the bit syntax. Unlike when constructing lists or tuples, the construction of a binary can fail with a badarg exception.

There can be zero or more segments in a binary to be constructed. The expression <<>> constructs a zero length binary.

Each segment in a binary can consist of zero or more bits. There are no alignment rules for individual segments of type integer and float. For binaries and bitstrings without size, the unit specifies the alignment. Since the default alignment for the binary type is 8, the size of a binary segment must be a multiple of 8 bits, that is, only whole bytes.

Example:

`<<Bin/binary,Bitstring/bitstring>>`

The variable Bin must contain a whole number of bytes, because the binary type defaults to unit:8. A badarg exception is generated if Bin consist of, for example, 17 bits.

The Bitstring variable can consist of any number of bits, for example, 0, 1, 8, 11, 17, 42, and so on. This is because the default unit for bitstrings is 1.

For clarity, it is recommended not to change the unit size for binaries. Instead, use binary when you need byte alignment and bitstring when you need bit alignment.

The following example successfully constructs a bitstring of 7 bits, provided that all of X and Y are integers:

`<<X:1,Y:6>>`

As mentioned earlier, segments have the following general syntax:

Value:Size/TypeSpecifierList

When constructing binaries, Value and Size can be any Erlang expression. However, for syntactical reasons, both Value and Size must be enclosed in parenthesis if the expression consists of anything more than a single literal or a variable. The following gives a compiler syntax error:

`<<X+1:8>>`

This expression must be rewritten into the following, to be accepted by the compiler:

`<<(X+1):8>>`

### Including Literal Strings

A literal string can be written instead of an element:

`<<"hello">>`

This is syntactic sugar for the following:

`<<$h,$e,$l,$l,$o>>`

## Matching Binaries

This section describes the rules for matching binaries, using the bit syntax.

There can be zero or more segments in a binary pattern. A binary pattern can occur wherever patterns are allowed, including inside other patterns. Binary patterns cannot be nested. The pattern <<>> matches a zero length binary.

Each segment in a binary can consist of zero or more bits. A segment of type binary must have a size evenly divisible by 8 (or divisible by the unit size, if the unit size has been changed). A segment of type bitstring has no restrictions on the size. A segment of type float must have size 64 or 32.

As mentioned earlier, segments have the following general syntax:

Value:Size/TypeSpecifierList

When matching Value, value must be either a variable or an integer, or a floating point literal. Expressions are not allowed.

Size must be a guard expression, which can use literals and previously bound variables. The following is not allowed:

`foo(N, <<X:N,T/binary>>) -> {X,T}.`

The two occurrences of N are not related. The compiler will complain that the N in the size field is unbound.

The correct way to write this example is as follows:

`foo(N, Bin) -> <<X:N,T/binary>> = Bin, {X,T}.`

Before OTP 23, Size was restricted to be an integer or a variable bound to an integer.

### Binding and Using a Size Variable

There is one exception to the rule that a variable that is as size must be previously bound. It is possible to match and bind a variable, and use it as a size within the the same binary pattern. For example:

`bar(<<Sz:8,Payload:Sz/binary-unit:8,Rest/binary>>) -> {Payload,Rest}.`

Here Sz is bound to the value in the first byte of the binary. Sz is then used at the number of bytes to match out as a binary.

Starting in OTP 23, the size can be a guard expression:

`bar(<<Sz:8,Payload:((Sz-1)*8)/binary,Rest/binary>>) -> {Payload,Rest}.`

Here Sz is the combined size of the header and the payload, so we will need to subtract one byte to get the size of the payload.

### Getting the Rest of the Binary or Bitstring

To match out the rest of a binary, specify a binary field without size:

`foo(<<A:8,Rest/binary>>) ->`

The size of the tail must be evenly divisible by 8.

To match out the rest of a bitstring, specify a field without size:

`foo(<<A:8,Rest/bitstring>>) ->`

There are no restrictions on the number of bits in the tail.

## Appending to a Binary

Appending to a binary in an efficient way can be done as follows:

`triples_to_bin(T) -> triples_to_bin(T, <<>>). triples_to_bin([{X,Y,Z} | T], Acc) -> triples_to_bin(T, <<Acc/binary,X:32,Y:32,Z:32>>); triples_to_bin([], Acc) -> Acc.`