Can I use member function as first argument to EnumWindows

Question

Can I use member function as first argument to EnumWindows? I don't see any workaround in this case even with boost::bind.

Or you can make your own `bind` function (with right calling convention) using `lParam` to pass it a pointer to member function... — Adriano Repetti, Dec 11 '13 at 14:05
@FrozenHeart, you should be able to use a lambda (I don't have access to a VS compiler at the moment). — hmjd, Dec 11 '13 at 14:08
@hmjd Capturing lambdas cannot be converted to free functions. — Raymond Chen, Dec 11 '13 at 14:15
@hmjd It's clear that capturing lambdas cannot be converted to free functions because how would it know where to get its captured variables from? — Raymond Chen, Dec 11 '13 at 14:29
@RaymondChen: trampoline. But that might be hard on exotic hardware. — MSalters, Dec 11 '13 at 14:33
@MSalters C++ has no provision for dynamic code generation. (Harvard architecture does not support it, for example. NX for another.) C++ is based on the principle that all code generation occurs at compile time. — Raymond Chen, Dec 11 '13 at 14:48

score 7 · Accepted Answer · edited May 23 '17 at 10:32

7

Given this normal callback function:

BOOL CALLBACK EnumWindowsProc(HWND wnd, LPARAM lParam);

You can invoke EnumWindows using lParam to pass it a pointer to your class:

EnumWindows(EnumWindowsProc, reinterpret_cast<LPARAM>(this));

In EnumWindowsProc you can simply call member function (casting lParam to your class type). Like this:

BOOL CALLBACK EnumWindowsProc(HWND wnd, LPARAM lParam)
{
    return reinterpret_cast<MyClass*>(lParam)->EnumWindows(wnd);
}

If you don't want to make your class method public you can:

Pack a struct to contain both class instance and pointer to member.
Use a library for delegates.
Use boost std:bind (in this case it'll work well because you do it on your own class member, it has not to be __stdcall).

Whatever you will use you can find more details in this post here on SO.

edited May 23 '17 at 10:32

Community

1
1

answered Dec 11 '13 at 14:12

Adriano Repetti

65,416
20
137
208

You are correct, I take back what I said. – John Bandela Dec 11 '13 at 14:20
Your compiler allow static_cast for 'this' to LPARAM? – manuell Dec 11 '13 at 14:32
@manuell pretty right, fixed...price to pay for uncompiled snippets wrote here! – Adriano Repetti Dec 11 '13 at 14:37
@Adriano I really would like to see the code for the std::bind case (bonus, it's in the OP question) – manuell Dec 11 '13 at 15:50
@manuell just look John's answer (or link in my answer to another SO post). It's little bit too prolix but more or less it's that. – Adriano Repetti Dec 11 '13 at 15:52
@Adriano That's using the LPARAM... Sigh... I hoped for some magic. Never mind. – manuell Dec 11 '13 at 15:58
@manuell yes, AFAIK no magic – Adriano Repetti Dec 11 '13 at 17:05

score 3 · Answer 2 · answered Dec 11 '13 at 14:18

EnumWindows takes a callback that looks like this

BOOL CALLBACK EnumWindowsProc(HWND hwnd,LPARAM lParam);

You cannot use either a member function or a lambda that has a capture. You do this instead.

typedef std::function<bool(HWND hwnd)> func_type;
BOOL CALLBACK MyEnumProc(HWND hwnd, LPARAM lparam)
{
    auto& func = *reinterpret_cast<func_type*>(lparam);
    if(func(hwnd))return TRUE;
    else return FALSE;

}

You would use it like this

auto f = std::make_shared<func_type>(std::bind(&mymemberfunction,this));
EnumWindows(MyEnumProc,reinterpret_cast<LPARAM>(f.get()));

//Make sure to keep the shared_ptr around until you are done enumerating

Can I use member function as first argument to EnumWindows

2 Answers2

Linked