January 10, 2012
If I want to have a method taking a callback function, I have to specify if it should take a function or delegate even if I don't really care. What's the best way to accept either? I cannot see any wrapper for something like this in std.typecons.


import std.stdio, std.traits;

void f(int i, void function(int) fn) {
    fn(i);
}

void d(int i, void delegate(int) dg) {
    dg(i);
}

// ugly..
void g(F)(int i, F callback) {
    static assert(isSomeFunction!F, "callback is not a function");
    static assert(__traits(compiles, { callback(i); }), "callback does not take int as a parameter");
    callback(i);
}

void fcb(int i) {
    writeln(i);
}

void main() {

    // Error: function t.f (int i, void function(int) fn) is not callable using argument types (int,void delegate(int))
    // f(1, (int i) { writeln(i); });
    f(2, &fcb);
    d(3, (int i) { writeln(i); });
    //Error: function t.d (int i, void delegate(int) dg) is not callable using argument types (int,void function(int))
    //d(4, &fcb);
    g(5, &fcb);
    g(6, (int i) { writeln(i); });
    //g(7, 7); // not a function
    //g(8, (string s) { }); // does not take int as arg
}

January 10, 2012
On 1/10/2012 10:05 PM, simendsjo wrote:
> If I want to have a method taking a callback function, I have to specify
> if it should take a function or delegate even if I don't really care.
> What's the best way to accept either? I cannot see any wrapper for
> something like this in std.typecons.

The simple way:

void callback(int i, void delegate(int) dg)
{
    dg(i);
}

void callback(int i, void function(int) fn)
{
    void wrap(int j)
    {
        function(j);
    }
    callback(i, &wrap);
}
January 10, 2012
On 1/10/2012 10:43 PM, Mike Parker wrote:
> On 1/10/2012 10:05 PM, simendsjo wrote:
>> If I want to have a method taking a callback function, I have to specify
>> if it should take a function or delegate even if I don't really care.
>> What's the best way to accept either? I cannot see any wrapper for
>> something like this in std.typecons.
>
> The simple way:
>
> void callback(int i, void delegate(int) dg)
> {
> dg(i);
> }
>
> void callback(int i, void function(int) fn)
> {
> void wrap(int j)
> {
> function(j);
> }
> callback(i, &wrap);
> }

And of course, wrap should be calling fn(j), not function(j)!
January 10, 2012
On 10.01.2012 14:43, Mike Parker wrote:
> On 1/10/2012 10:05 PM, simendsjo wrote:
>> If I want to have a method taking a callback function, I have to specify
>> if it should take a function or delegate even if I don't really care.
>> What's the best way to accept either? I cannot see any wrapper for
>> something like this in std.typecons.
>
> The simple way:
>
> void callback(int i, void delegate(int) dg)
> {
> dg(i);
> }
>
> void callback(int i, void function(int) fn)
> {
> void wrap(int j)
> {
> function(j);
> }
> callback(i, &wrap);
> }

Yeah, but a bit tedious.. I found toDelegate: http://dlang.org/phobos/std_functional.html#toDelegate
January 10, 2012
On 2012-01-10 14:48, simendsjo wrote:
> On 10.01.2012 14:43, Mike Parker wrote:
>> On 1/10/2012 10:05 PM, simendsjo wrote:
>>> If I want to have a method taking a callback function, I have to specify
>>> if it should take a function or delegate even if I don't really care.
>>> What's the best way to accept either? I cannot see any wrapper for
>>> something like this in std.typecons.
>>
>> The simple way:
>>
>> void callback(int i, void delegate(int) dg)
>> {
>> dg(i);
>> }
>>
>> void callback(int i, void function(int) fn)
>> {
>> void wrap(int j)
>> {
>> function(j);
>> }
>> callback(i, &wrap);
>> }
>
> Yeah, but a bit tedious.. I found toDelegate:
> http://dlang.org/phobos/std_functional.html#toDelegate

Or make it a template parameter and check if it's callable using std.traits.isCallable.

-- 
/Jacob Carlborg
January 10, 2012
On 10.01.2012 14:43, Mike Parker wrote:
> On 1/10/2012 10:05 PM, simendsjo wrote:
>> If I want to have a method taking a callback function, I have to specify
>> if it should take a function or delegate even if I don't really care.
>> What's the best way to accept either? I cannot see any wrapper for
>> something like this in std.typecons.
>
> The simple way:
>
> void callback(int i, void delegate(int) dg)
> {
> dg(i);
> }
>
> void callback(int i, void function(int) fn)
> {
> void wrap(int j)
> {
> function(j);
> }
> callback(i, &wrap);
> }

I tried the following, but I get some error messages:

h(9, (int i) { writeln(i); });

t.d(46): Error: template t.h(F) if (isCompatibleFunction!(F,void function(int))) does not match any function template declaration
t.d(46): Error: template t.h(F) if (isCompatibleFunction!(F,void function(int))) cannot deduce template function from argument types !()(int,void delegate(int))


template isCompatibleFunction(Src, Dest) {
    static assert(isSomeFunction!Src, "Source is not a function");
    static assert(isSomeFunction!Dest, "Destination is not a function");
    enum bool isCompatibleFunction =
        is(ParameterTypeTuple!Src == ParameterTypeTuple!Dest) &&
        is(ParameterStorageClassTuple!Src == ParameterStorageClassTuple!Dest) &&
        is(ReturnType!Src == ReturnType!Dest);
}

void h(F)(int i, F callback) if(isCompatibleFunction!(F, void function(int))) {
    callback(i);
}

January 10, 2012
On 10.01.2012 15:53, Jacob Carlborg wrote:
> On 2012-01-10 14:48, simendsjo wrote:
>> On 10.01.2012 14:43, Mike Parker wrote:
>>> On 1/10/2012 10:05 PM, simendsjo wrote:
>>>> If I want to have a method taking a callback function, I have to
>>>> specify
>>>> if it should take a function or delegate even if I don't really care.
>>>> What's the best way to accept either? I cannot see any wrapper for
>>>> something like this in std.typecons.
>>>
>>> The simple way:
>>>
>>> void callback(int i, void delegate(int) dg)
>>> {
>>> dg(i);
>>> }
>>>
>>> void callback(int i, void function(int) fn)
>>> {
>>> void wrap(int j)
>>> {
>>> function(j);
>>> }
>>> callback(i, &wrap);
>>> }
>>
>> Yeah, but a bit tedious.. I found toDelegate:
>> http://dlang.org/phobos/std_functional.html#toDelegate
>
> Or make it a template parameter and check if it's callable using
> std.traits.isCallable.
>

Like this?
void callback(F)(int i, F fn) if(isCallable!F) {
  fn(i);
}

.. but then the parameters wouldn't be documented.
January 10, 2012
On 01/10/2012 06:53 AM, Jacob Carlborg wrote:
> On 2012-01-10 14:48, simendsjo wrote:
>> On 10.01.2012 14:43, Mike Parker wrote:
>>> On 1/10/2012 10:05 PM, simendsjo wrote:
>>>> If I want to have a method taking a callback function, I have to
>>>> specify
>>>> if it should take a function or delegate even if I don't really care.
>>>> What's the best way to accept either? I cannot see any wrapper for
>>>> something like this in std.typecons.
>>>
>>> The simple way:
>>>
>>> void callback(int i, void delegate(int) dg)
>>> {
>>> dg(i);
>>> }
>>>
>>> void callback(int i, void function(int) fn)
>>> {
>>> void wrap(int j)
>>> {
>>> function(j);
>>> }
>>> callback(i, &wrap);
>>> }
>>
>> Yeah, but a bit tedious.. I found toDelegate:
>> http://dlang.org/phobos/std_functional.html#toDelegate
>
> Or make it a template parameter and check if it's callable using
> std.traits.isCallable.
>
What's wrong with toDelegate ? Seems to be pretty handy.

//simple snip
import std.functional;

int main()
{
    int delegate( int i) dg;
    alias dg callback;
    callback = toDelegate(&test);
    writeln( callback( 12 ) );
    readln();

    return 0;
}

int test(int i) { return 30 +i;}


January 10, 2012
On 2012-01-10 20:24, bls wrote:
> On 01/10/2012 06:53 AM, Jacob Carlborg wrote:
>> On 2012-01-10 14:48, simendsjo wrote:
>>> On 10.01.2012 14:43, Mike Parker wrote:
>>>> On 1/10/2012 10:05 PM, simendsjo wrote:
>>>>> If I want to have a method taking a callback function, I have to
>>>>> specify
>>>>> if it should take a function or delegate even if I don't really care.
>>>>> What's the best way to accept either? I cannot see any wrapper for
>>>>> something like this in std.typecons.
>>>>
>>>> The simple way:
>>>>
>>>> void callback(int i, void delegate(int) dg)
>>>> {
>>>> dg(i);
>>>> }
>>>>
>>>> void callback(int i, void function(int) fn)
>>>> {
>>>> void wrap(int j)
>>>> {
>>>> function(j);
>>>> }
>>>> callback(i, &wrap);
>>>> }
>>>
>>> Yeah, but a bit tedious.. I found toDelegate:
>>> http://dlang.org/phobos/std_functional.html#toDelegate
>>
>> Or make it a template parameter and check if it's callable using
>> std.traits.isCallable.
>>
> What's wrong with toDelegate ? Seems to be pretty handy.
>
> //simple snip
> import std.functional;
>
> int main()
> {
> int delegate( int i) dg;
> alias dg callback;
> callback = toDelegate(&test);
> writeln( callback( 12 ) );
> readln();
>
> return 0;
> }
>
> int test(int i) { return 30 +i;}

A template parameter with a template constraint will accept any callable type. Function pointer, delegate, struct/class overloading the call operator and so on.

-- 
/Jacob Carlborg
January 11, 2012
On 10/01/2012 19:56, Jacob Carlborg wrote:
<snip>
> A template parameter with a template constraint will accept any callable type. Function
> pointer, delegate, struct/class overloading the call operator and so on.

Indeed, this is done in the C++ STL quite a lot.

The drawback is that templated methods lose their virtuality, because it cannot be known in advance on what types the template will be instantiated in order to populate the vtable.

FWIW my utility library includes a delegate wrapper:
http://pr.stewartsplace.org.uk/d/sutil/

(dgwrap works in both D1 and D2, though other bits of the library need updating to current D2)

Stewart.
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