two variables when defining function

Question

I'm new to C and I understand that you define a function using this format

return_type function_name(args)
{
    ... body ...
}

where a function definition looks like this:

void __init tick_broadcast_init(void)
{
    ... body ...
}

It looks like there's two variables, __init and tick_broadcast_init. Anyone know what's going on?

Between the return type and the function name there may be compiler-specific attributes (such as directives governing the section of the executable, the calling convention and other stuff). This is the case for `__init` (actually a macro which expands to several of these attributes). — Matteo Italia, Oct 19 '15 at 00:17

coincoin · Accepted Answer · 2015-10-19T01:16:15.310

2

__init and tick_broadcast_init are not variables.

__init is a a macro see here for more explanations.

tick_broadcast_init is your function name.

Your return type is void meaning "nothing" so that there is in fact no data in return.

edited Oct 19 '15 at 01:16

answered Oct 19 '15 at 00:16

coincoin

1 Answers1