alias int function(int) function_type;



void main()
{
	bar!foo(2);
	bar!((int i)=> i*2)(2);
}

int foo(int i)
{
    return i;
}

void bar(alias baz)(int i)
{
	static if (!is(typeof(baz) == function_type))
	{
		pragma(msg, typeof(baz), " != ", function_type); //wtf?
	}

	std.stdio.writeln(bar(i));
}

Where am I wrong?

On Tuesday, 1 October 2013 at 18:07:31 UTC, Roman wrote:
> alias int function(int) function_type;
>
>
>
> void main()
> {
> 	bar!foo(2);
> 	bar!((int i)=> i*2)(2);
> }
>
> int foo(int i)
> {
>     return i;
> }
>
> void bar(alias baz)(int i)
> {
> 	static if (!is(typeof(baz) == function_type))
> 	{
> 		pragma(msg, typeof(baz), " != ", function_type); //wtf?
> 	}
>
> 	std.stdio.writeln(bar(i));
> }
>
> Where am I wrong?

bar!foo should work as you expect.
bar!((int i)=> i*2) is different because (int i) => i*2 is probably a template. Can you paste the output of your code so we can help you more easily ?

On Tuesday, 1 October 2013 at 18:10:05 UTC, deadalnix wrote:
> bar!((int i)=> i*2) is different because (int i) => i*2 is probably a template.

Also, it will be inferred as pure, nothrow and @safe.

David

> Can you paste the output of your code so we can help you more easily ?

int(int i) != int function(int)
int function(int i) pure nothrow @safe != int function(int)
2
4

So, why 1st case incorrect?

On Tuesday, 1 October 2013 at 18:38:33 UTC, Roman wrote:
>> Can you paste the output of your code so we can help you more easily ?
>
> int(int i) != int function(int)
> int function(int i) pure nothrow @safe != int function(int)
> 2
> 4
>
> So, why 1st case incorrect?

The first case is a bug to me. Can you fill it in http://d.puremagic.com/issues/ please ? The second is expected.

What you should do is test via is(typeof(...)) : function_type). The : test for implicit conversion, and int function(int i) pure nothrow @safe convert implicitly to int function(int) .

On Tuesday, 1 October 2013 at 18:07:31 UTC, Roman wrote:
> alias int function(int) function_type;
>
>

function_type is a function pointer

> void main()
> {
> 	bar!foo(2);

foo is a function, not a function pointer

> 	bar!((int i)=> i*2)(2);
> }
>
> int foo(int i)
> {
>     return i;
> }
>
> void bar(alias baz)(int i)
> {
> 	static if (!is(typeof(baz) == function_type))
> 	{
> 		pragma(msg, typeof(baz), " != ", function_type); //wtf?
> 	}

here you are testing that baz is a function pointer

> 	std.stdio.writeln(bar(i));

you probably meant baz here

> }
>
> Where am I wrong?

Wrong at place where you mix function type and function pointer. You are facing

int(int i) != int function(int)
int function(int i) pure nothrow @safe != int function(int)

On Tuesday, 1 October 2013 at 18:10:05 UTC, deadalnix wrote:
> On Tuesday, 1 October 2013 at 18:07:31 UTC, Roman wrote:
>> alias int function(int) function_type;
>>
>>
>>
>> void main()
>> {
>> 	bar!foo(2);
>> 	bar!((int i)=> i*2)(2);
>> }
>>
>> int foo(int i)
>> {
>>    return i;
>> }
>>
>> void bar(alias baz)(int i)
>> {
>> 	static if (!is(typeof(baz) == function_type))
>> 	{
>> 		pragma(msg, typeof(baz), " != ", function_type); //wtf?
>> 	}
>>
>> 	std.stdio.writeln(bar(i));
>> }
>>
>> Where am I wrong?
>
> bar!foo should work as you expect.
> bar!((int i)=> i*2) is different because (int i) => i*2 is probably a template. Can you paste the output of your code so we can help you more easily ?

???

foo is not a function pointer

auto x = &foo;
bar!(x)(2);

should work.

And (int i) => i*2 is of course not a template [(i) => i*2 would be].

On Tuesday, 1 October 2013 at 18:46:45 UTC, deadalnix wrote:
> On Tuesday, 1 October 2013 at 18:38:33 UTC, Roman wrote:
>>> Can you paste the output of your code so we can help you more easily ?
>>
>> int(int i) != int function(int)
>> int function(int i) pure nothrow @safe != int function(int)
>> 2
>> 4
>>
>> So, why 1st case incorrect?
>
> The first case is a bug to me. Can you fill it in http://d.puremagic.com/issues/ please ? The second is expected.
>
> What you should do is test via is(typeof(...)) : function_type). The : test for implicit conversion, and int function(int i) pure nothrow @safe convert implicitly to int function(int) .

With ":" bar!((int i)=> i*2)(2) works properly, but bar!foo(2) still doesn't pass expression ( evidently foo doesn't implicitly convert to function_type)

> you probably meant baz here
yes

> auto x = &foo;
> bar!(x)(2);

> should work.

Yeah, it worked! Thanks!

On Tuesday, 1 October 2013 at 19:11:26 UTC, Roman wrote:
> On Tuesday, 1 October 2013 at 18:46:45 UTC, deadalnix wrote:
>> On Tuesday, 1 October 2013 at 18:38:33 UTC, Roman wrote:
>>>> Can you paste the output of your code so we can help you more easily ?
>>>
>>> int(int i) != int function(int)
>>> int function(int i) pure nothrow @safe != int function(int)
>>> 2
>>> 4
>>>
>>> So, why 1st case incorrect?
>>
>> The first case is a bug to me. Can you fill it in http://d.puremagic.com/issues/ please ? The second is expected.
>>
>> What you should do is test via is(typeof(...)) : function_type). The : test for implicit conversion, and int function(int i) pure nothrow @safe convert implicitly to int function(int) .
>
> With ":" bar!((int i)=> i*2)(2) works properly, but bar!foo(2) still doesn't pass expression ( evidently foo doesn't implicitly convert to function_type)

You are not in C, where 'function_name' (except some circumstances) is implicitly converted to pointer to that function.