Re: Call a function with a function pointer

[Benjamin Thaut]

Am 10.10.2013 17:45, schrieb Namespace:
On Thursday, 10 October 2013 at 15:15:45 UTC, bearophile wrote:
Namespace:
You mean like this?
----
void foo(T)(extern(C) void function(T*) func) {

}
----

That prints: Error: basic type expected, not extern

In theory that's correct, in practice the compiler refuses that, it's
in Bugzilla, so try to define the type outside the signature (untested):

alias TF = extern(C) void function(T*);

void foo(T)(TF func) {}

Bye,
bearophile

/d917/f732.d(8): Error: basic type expected, not extern
/d917/f732.d(8): Error: semicolon expected to close alias declaration
/d917/f732.d(8): Error: no identifier for declarator void function(T*)

I found a possible workaround. Its ugly as hell, but at least it works 
until the bugs are fixed. The trick is to make a helper struct. Define 
the function you want within that, and then use typeof to get the type.


import std.stdio;

extern(C) void testFunc(int* ptr)
{
        *ptr = 5;
}

struct TypeHelper(T)
{
        extern(C) static void func(T*);
        alias typeof(&func) func_t;
}

void Foo(T)(TypeHelper!T.func_t func, T* val)
{
        func(val);
}

void main(string[] args)
{
        pragma(msg, TypeHelper!int.func_t.stringof);
        int test = 0;
        Foo!int(&testFunc, &test);
        writefln("%d", test);
}

--
Kind Regards
Benjamin Thaut