From: ts <decoux@...>
Date: 2001-07-20T23:23:14+09:00
Subject: [ruby-talk:18187] Re: Invoking a method from C side

>>>>> "L" == Lothar Scholz <llothar@mailandnews.de> writes:

L> The "method_call" function is declared private. So how can i
L> make a  " methodobj.call("foo")  " call from the C-Side without
L> to much overhead.

 rb_funcall(methodobj, rb_intern("call"), 1, rb_str_new2("foo"));

L> Isn't it possible to make "method_call"  and the rb_cMethod variable
L> public  (Of course method_call should be renamed to rb_method_call") ?

 Try this

pigeon% grep \"call\" *.c
eval.c: return rb_funcall2(cmd, rb_intern("call"),
eval.c:             call_trace_func("call", b2->nd_file, nd_line(b2),
eval.c:    rb_define_method(rb_cProc, "call", proc_call, -2);
eval.c:    rb_define_method(rb_cMethod, "call", method_call, -1);
eval.c:    rb_define_method(rb_cUnboundMethod, "call", umethod_call, -1);
eval.c:    rb_define_method(rb_cCont, "call", rb_cont_call, -1);
marshal.c:      rb_funcall(arg->proc, rb_intern("call"), 1, v);
pigeon% 


 #call is defined for Proc, Method, UnboundMethod and Continuation. This
  mean that in your C program you'll end by making the test for all
  possibles calls.


Guy Decoux