[erlang-questions] Accessing A Binary as an Array of Floats/Doubles Within a NIF
Sverker Eriksson
sverker.eriksson@REDACTED
Mon Jun 20 21:50:58 CEST 2016
If you use <<1.0/float-native>>
then it's just type cast to a 'double' pointer:
double* my_array;
my_array = (double*) Bin.data;
printf("%lf", my_array[0]);
Default byte order is 'float-big', which will not work on little endian
machines (x86).
Depending on architecture you also might have to ensure 8 byte alignment.
x86 will accept unaligned floats but the performance may suffer.
/Sverker, Erlang/OTP
On 06/20/2016 09:04 PM, John Duffy wrote:
> Hi
> I'm using FreeBSD 10.3 & Erlang 18.3 and I'm trying to get to grips with passing a binary of floats to a NIF for further processing as as array of C doubles.
> My simple NIF module compiles and I can access my "my_nif_1" function from the erlang shell. I can access the 'size' structure member but I can't figure out how to access the 'data' structure member as a C array. I understand that 'data' is a block of memory of UINT bytes but I can't figure out how to "cast" this to a C array within the NIF.
> Any help would be mud appreciated.
> Kind regards
> John
> #include "erl_nif.h"
> static ERL_NIF_TERM my_nif_1(ErlNifEnv * env, int argc, const ERL_NIF_TERM argv[])
> {
> ErlNifBinary Bin;
> if (!enif_inspect_binary(env, argv[0], &Bin)) {
> return enif_make_badarg(env);
> }
> // So far, so good...
> // From the erl shell...
> // my_nifs:my_nif_1(<<1.0/float>>). returns a Bin.size of 8
> // my_nifs:my_nif_1(<<1.0/float,2.0/float>>. returns a Bin.size of 16
> //
> // How do I access Bin.data as an array of doubles?
> //
> return enif_make_int(env, Bin.size); // For testing... this works.
> }
> static ErlNifFunc nif_funcs[] = {
> {"my_nif_1", 1, my_nif_1}
> };
> ERL_NIF_INIT(my_nifs, nif_funcs, NULL, NULL, NULL, NULL);
>
>
>
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