CODEWITHSUNDEEP
erlang
why
why

Reputation: 24851

How to define binary in another binary?

A = <<"hello">>.
B = <<A:80/binary, 100:8>>.

It gives me:

** exception error: bad argument

and <<"hello">>. works, but:

A = "hello".
<<A>>.

can not work.

Upvotes: 2

Views: 154

Answers (2)

Hynek -Pichi- Vychodil
Hynek -Pichi- Vychodil

Reputation: 26121

A doesn't have size 80 bytes which obviously doesn't match A:80/binary in first case. 

1> A = <<"hello">>.
<<"hello">>
2> B = <<A/binary, 100:8>>.
<<"hellod">>
3> Pad = 80 - size(A), C = <<A/binary, 0:Pad/unit:8, 100:8>>.
<<104,101,108,108,111,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
  0,0,0,0,0,0,...>>

<<"hello">> is syntactic sugar for <<$h,$e,$l,$l,$o>>. Bit syntax expression assumes 8/integer,unsigned,big,unit:1 type specification by default. A is not integer so <<A>> raises badarg exception in second case.

Upvotes: 4

I GIVE CRAP ANSWERS
I GIVE CRAP ANSWERS

Reputation: 18869

The value <<"Hello">> works but only because "Hello" is a string literal. When you write,

  A = "Hello",

you are creating a String object, which is really a list of unicode codepoints. Now, when you declare,

  <<A>>

then A is assumed to be an integer because the is the default. Naturally something is wrong when you try to inject a list/string for an integer, which is the reason for the badarg.

The solution is two-fold:

  list_to_binary(A)

will convert the list to a binary. Now you have the equivalent of <<A/binary>> and you can manipulate it:

   L = byte_size(A),
   <<L:32/integer, A/binary>>

Upvotes: 0

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