is there a way to have midl generate each interface in a separate .h ?

Question

I have a bunch of objects that inherit abstracts interfaces generated from an idl file. Each object that use of theses interfaces include the same file interfaces.h which contain all the c++ generated abstract classes that map to the idl interface.

Each time I change anything into interfaces.idl every classes that depend on this have to be rebuild since interfaces.h change. Is there a flag or something to tell midl to generate each abstract class in its own .h ?

Answer 1

The only way I can think of is to put each interface in its own IDL file, or divide them into multiple IDLs according to rate-of-change.

Then include (or is it #import -- I forget) these interface IDLs into the main library IDL, which will produce the type library, if you need that.

Answer 2

Here is a sample on how to organize the idl to generate separate .h files and a single typelib. The correct directive is import.

main.idl

import "oaidl.idl";
import "ocidl.idl";

import "frag1.idl";
import "frag2.idl";

[
    uuid(1BECE2AF-2792-49b9-B133-BBC89C850D6F),
    version(1.0),
    helpstring("Bibliothèque de types")
]
library Library
{
    importlib("stdole2.tlb");

    interface IFrag1;
    interface IFrag2;
}

frag1.idl

import "oaidl.idl";
import "ocidl.idl";

[
    object,
    uuid(9AEB517B-48B9-4628-8DD3-4A0BA8D39BEF),
    dual,
    nonextensible,
    helpstring("Interface IFrag1"),
    pointer_default(unique)
]
interface IFrag1 : IDispatch {
    HRESULT frag1();
};

frag2.idl

import "oaidl.idl";
import "ocidl.idl";

[
    object,
    uuid(D60835D4-E1B1-40fb-B583-A75373EF15BE),
    dual,
    nonextensible,
    helpstring("Interface IFrag2"),
    pointer_default(unique)
]
interface IFrag2 : IDispatch {
    HRESULT frag2();
};