digitalmars.D - why are types mismatch?

Roman (19/19) Oct 01 2013 alias int function(int) function_type;

deadalnix (5/24) Oct 01 2013 bar!foo should work as you expect.

David Nadlinger (3/5) Oct 01 2013 Also, it will be inferred as pure, nothrow and @safe.

Roman (5/7) Oct 01 2013 int(int i) != int function(int)

deadalnix (6/13) Oct 01 2013 The first case is a bug to me. Can you fill it in

Roman (4/20) Oct 01 2013 With ":" bar!((int i)=> i*2)(2) works properly, but bar!foo(2)

Roman (2/6) Oct 01 2013 Yeah, it worked! Thanks!
Maxim Fomin (4/26) Oct 01 2013 You are not in C, where 'function_name' (except some

Maxim Fomin (8/39) Oct 01 2013 ???

deadalnix (4/5) Oct 01 2013 Hoooo ! I see. Well, it should. This distinction between function

Maxim Fomin (9/28) Oct 01 2013 foo is a function, not a function pointer

"Roman" <shamyan.roman gmail.com> writes:

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?

Oct 01 2013

"deadalnix" <deadalnix gmail.com> writes:

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 ?

Oct 01 2013

"David Nadlinger" <code klickverbot.at> writes:

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

Oct 01 2013

"Roman" <shamyan.roman gmail.com> writes:

 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?

Oct 01 2013

"deadalnix" <deadalnix gmail.com> writes:

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) .

Oct 01 2013

"Roman" <shamyan.roman gmail.com> writes:

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)

Oct 01 2013

"Roman" <shamyan.roman gmail.com> writes:

 you probably meant baz here

yes

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

 should work.

Yeah, it worked! Thanks!

Oct 01 2013

"Maxim Fomin" <maxim maxim-fomin.ru> writes:

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.

Oct 01 2013

"Maxim Fomin" <maxim maxim-fomin.ru> writes:

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].

Oct 01 2013

"deadalnix" <deadalnix gmail.com> writes:

On Tuesday, 1 October 2013 at 18:55:42 UTC, Maxim Fomin wrote:
 foo is not a function pointer

Hoooo ! I see. Well, it should. This distinction between function 
pointer and function do not even exists at assembly level (you 
always pass a function pointer to call instruction).

Oct 01 2013

"Maxim Fomin" <maxim maxim-fomin.ru> writes:

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)

Oct 01 2013

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digitalmars.D - why are types mismatch?