I need to generate a variable which has the following properties - 32 bit, big-endian integer, initialized with 0x00000001 (I'm going to increment that number one by one). Is there a syntax in erlang for this?
In Erlang, normally you'd keep such numbers as plain integers inside the program:
X = 1.
or equivalently, if you want to use a hexadecimal literal:
X = 16#00000001.
And when it's time to convert the number to a binary representation in order to send it somewhere else, use bit syntax:
<<X:32/big>>
This returns a binary containing four bytes:
<<0,0,0,1>>
(That's a 32-bit big-endian integer. In fact, big-endian is the default, so you could just write <<X:32>>
. <<X:64/little>>
would be a 64-bit little-endian integer.)
On the other hand, if you just want to print the number in 0x00000001
format, use io:format
with this format specifier:
io:format("0x~8.16.0b~n", [X]).
The 8
tells it to use a field width of 8 characters, the 16
tells it to use radix 16 (i.e. hexadecimal), and the 0
is the padding character, used for filling the number up to the field width.
Note that incrementing a variable works differently in Erlang compared to other languages. Once a variable has been assigned a value, you can't change it, so you'd end up making a recursive call, passing the new value as an argument to the function. This answer has an example.
According to the documentation[1] the following snippet should generate a 32-bit signed integer in little endian.
1> I = 258.
258
2> B = <<I:4/little-signed-integer-unit:8>>.
<<2,1,0,0>>
And the following should produce big endian numbers:
1> I = 258.
258
2> B = <<I:4/big-signed-integer-unit:8>>.
<<0,0,1,2>>
[1] http://erlang.org/doc/programming_examples/bit_syntax.html
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