Re: Call a function with a function pointer
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
Re: Call a function with a function pointer
On 10/13/13 16:43, Benjamin Thaut wrote:
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.
There's no need for such ugly workarounds -- this is just a problem with
the *new* alias syntax. The old one accepts it (unless this changed recently):
alias extern(C) static void function(int*) Func_t;
artur
Re: Call a function with a function pointer
Am 13.10.2013 17:17, schrieb Artur Skawina:
On 10/13/13 16:43, Benjamin Thaut wrote:
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.
There's no need for such ugly workarounds -- this is just a problem with
the *new* alias syntax. The old one accepts it (unless this changed recently):
alias extern(C) static void function(int*) Func_t;
artur
Oh so this bug was fixed? Thats good to know.
Re: Call a function with a function pointer
On 10/10/13 20:54, Dicebot wrote:
On Thursday, 10 October 2013 at 17:47:54 UTC, Namespace wrote:
import std.stdio;
void foo1(void function(void*) fp) { }
void foo2(void function(int) fp) { }
void foo3(void*) { }
void main()
{
foo1((void* ptr) = ( assert(ptr is null) ));
foo2((int a) = ( a + 1 )); /// Fails: Error: function foo2 (void
function(int) fp) is not callable using argument types (int function(int a)
pure nothrow @safe)
foo1(foo3);
void foo4(void function(void*) fp) { }
foo1(foo4); /// Fails: Error: function foo1 (void function(void*) fp)
is not callable using argument types (void delegate(void function(void*) fp))
}
You are using short lambda syntax a = b. Here `b` is always return
statement. It is equivalent to (a) { return b; }. And your `foo2` signature
expects lambda returning void, like (a) { return; }
Second error is DMD incompetence in deducing minimal required type of nested
function. It always treats them as delegates (== having hidden context
pointer) even if those do not refer any actual context. And plain lambdas are
of course binary incompatible with delegates (closures) because of that extra
pointer field.
It's probably not just incompetence (the compiler is able to figure this
out in other contexts), but a deliberate choice. Having function types
depend on their bodies would not be a good idea. Eg
int c;
auto f() {
int a = 42;
int f1() { return a; }
int f2() { return 0; }
return !c?f1:f2;
}
Mark f2 as 'static' and this code will no longer compile. If that would
be done automatically then you'd have to 'undo' it manually, which would
cause even more problems (consider generic code, which isn't prepared
to handle this).
artur
[1] at least without other language improvements; enabling overloading on
'static', plus a few other enhancements, would change the picture.
Re: Call a function with a function pointer
On Friday, 11 October 2013 at 15:55:17 UTC, Artur Skawina wrote:
It's probably not just incompetence (the compiler is able to
figure this
out in other contexts), but a deliberate choice. Having
function types
depend on their bodies would not be a good idea. Eg
int c;
auto f() {
int a = 42;
int f1() { return a; }
int f2() { return 0; }
return !c?f1:f2;
}
Mark f2 as 'static' and this code will no longer compile. If
that would
be done automatically then you'd have to 'undo' it manually,
which would
cause even more problems (consider generic code, which isn't
prepared
to handle this).
artur
[1] at least without other language improvements; enabling
overloading on
'static', plus a few other enhancements, would change the
picture.
Agreed.
However, I do feel uncomfortable with new habit to put `static`
everywhere to avoid hidden compiler help :(
Re: Call a function with a function pointer
On Thursday, 10 October 2013 at 14:13:47 UTC, Namespace wrote:
I have this function:
void foo(T)(void function(T*) test) { }
And want to call it with a C function:
foo!(SDL_Surface)(SDL_FreeSurface);
but I get:
Fehler 1 Error: foo (void function(SDL_Surface*) test) is not
callable using argument types (extern (C) void
function(SDL_Surface*) nothrow)
What would be the smartest solution?
Wrap it in a lambda. Or change foo() signature to accept
`extern(C)` functions - you can't just mix calling convention.
Re: Call a function with a function pointer
Am 10.10.2013 16:13, schrieb Namespace:
I have this function:
void foo(T)(void function(T*) test) { }
And want to call it with a C function:
foo!(SDL_Surface)(SDL_FreeSurface);
but I get:
Fehler1Error: foo (void function(SDL_Surface*) test) is not
callable using argument types (extern (C) void function(SDL_Surface*)
nothrow)
What would be the smartest solution?
If you can change the signature of foo just add a extern(c) to the
function pointer declaration.
Otherwise just wrap the SDL_FreeSurface call into a delegate on the
caller side.
--
Kind Regards
Benjamin Thaut
Re: Call a function with a function pointer
On Thursday, 10 October 2013 at 14:40:09 UTC, Namespace wrote:
Example? I do not use lambdas often.
void foo(T)(void function(T*) test)
{
}
extern(C) void bar(int*) { }
void main()
{
foo( (int* a) = bar(a) );
}
I don't know to what extent IFTI can work here though.
Re: Call a function with a function pointer
On Thursday, 10 October 2013 at 14:28:20 UTC, Dicebot wrote:
On Thursday, 10 October 2013 at 14:13:47 UTC, Namespace wrote:
I have this function:
void foo(T)(void function(T*) test) { }
And want to call it with a C function:
foo!(SDL_Surface)(SDL_FreeSurface);
but I get:
Fehler 1 Error: foo (void function(SDL_Surface*) test) is not
callable using argument types (extern (C) void
function(SDL_Surface*) nothrow)
What would be the smartest solution?
Wrap it in a lambda.
Example? I do not use lambdas often.
Re: Call a function with a function pointer
On Thursday, 10 October 2013 at 14:44:00 UTC, Dicebot wrote:
On Thursday, 10 October 2013 at 14:40:09 UTC, Namespace wrote:
Example? I do not use lambdas often.
void foo(T)(void function(T*) test)
{
}
extern(C) void bar(int*) { }
void main()
{
foo( (int* a) = bar(a) );
}
I don't know to what extent IFTI can work here though.
That works. Thanks.
Re: Call a function with a function pointer
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*)
Re: Call a function with a function pointer
Namespace:
/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*)
It seems that even the new alias syntax doesn't support the
extern :-) Perhaps this bug is not yet in Bugzilla.
Try:
alias extern(C) void function(T*) TF;
void foo(T)(TF func) {}
Bye,
bearophile
Re: Call a function with a function pointer
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
That is limitation of current extern - it can only be attached to
symbol declarations, not types. AFAIK you need to do `extern(C)
alias TF = ...` but anyway this method is very likely to break
IFTI completely.
Re: Call a function with a function pointer
import std.stdio;
void foo1(void function(void*) fp) { }
void foo2(void function(int) fp) { }
void foo3(void*) { }
void main()
{
foo1((void* ptr) = ( assert(ptr is null) ));
foo2((int a) = ( a + 1 )); /// Fails: Error: function foo2
(void function(int) fp) is not callable using argument types (int
function(int a) pure nothrow @safe)
foo1(foo3);
void foo4(void function(void*) fp) { }
foo1(foo4); /// Fails: Error: function foo1 (void
function(void*) fp) is not callable using argument types (void
delegate(void function(void*) fp))
}
Can someone explain that to me?
Re: Call a function with a function pointer
On Thursday, 10 October 2013 at 17:47:54 UTC, Namespace wrote:
import std.stdio;
void foo1(void function(void*) fp) { }
void foo2(void function(int) fp) { }
void foo3(void*) { }
void main()
{
foo1((void* ptr) = ( assert(ptr is null) ));
foo2((int a) = ( a + 1 )); /// Fails: Error: function foo2
(void function(int) fp) is not callable using argument types
(int function(int a) pure nothrow @safe)
foo1(foo3);
void foo4(void function(void*) fp) { }
foo1(foo4); /// Fails: Error: function foo1 (void
function(void*) fp) is not callable using argument types (void
delegate(void function(void*) fp))
}
Can someone explain that to me?
You are using short lambda syntax a = b. Here `b` is always
return statement. It is equivalent to (a) { return b; }. And
your `foo2` signature expects lambda returning void, like (a) {
return; }
Second error is DMD incompetence in deducing minimal required
type of nested function. It always treats them as delegates (==
having hidden context pointer) even if those do not refer any
actual context. And plain lambdas are of course binary
incompatible with delegates (closures) because of that extra
pointer field.
Re: Call a function with a function pointer
On 10/10/13, bearophile bearophileh...@lycos.com wrote: Perhaps this bug is not yet in Bugzilla. I'm pretty sure I saw it filed somewhere. Can't find it though..
Re: Call a function with a function pointer
Andrej Mitrovic: I'm pretty sure I saw it filed somewhere. Can't find it though.. I have just added the new test case :-) http://d.puremagic.com/issues/show_bug.cgi?id=6754 Bye, bearophile
